[The quadratic formula][section 67]

(1.) x² - 4x = 21

-21 -21 subtract 21 from each side
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x² - 4x - 21 = 0
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x = [-b ± √b² - 4ac]/(2a) use the quadratic formula
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x = [4 ± √4² - 4*1*-21]/(2*1) make substitutions

x = [4 ± 10]/2 multiply add take square root

x = 7 x = -3 add subtract divide

results: x = 7 , x = -3

(2.) x² + 6x = 16 here is the problem

-16 -16 subtract 16 from each side
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x² + 6x - 16 = 0 subtract
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x = [-b ± √b² - 4ac]/(2a) use the quadratic formula
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x = [-6 ± √6² - 4*1*-16]/(2*1) make substitutions

x = [-6 ± 10]/2 multiply add and take sq root

x = 2 x = -8 add subtract divide

results: x = 2 , x = -8

(3.) 2x² - x - 3 = 0 here is the problem
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x = [-b ± √b² - 4ac]/(2a) use the quadratic formula
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x = [1 ± √1² - 4*2*-3]/(2*2) make substitutions

x = (1 ± 5)/4 multiply add take square root

x = 3/2 x = -1 add subtract and divide

results: x = 3/2 , x = -1

(4.) a² + 4a = 5 here is the problem

-5 -5 subtract 5 from each side
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a² + 4a - 5 = 0 subtract
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a = [-b ± √b² - 4ac]/(2a) use the quadratic formula
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a = [-4 ± √4² - 4*1*-5]/(2*1) make substitutions

a = [-4 ± 6]/2 multiply add & take sq root

a = 1 a = -5 add subtract and divide

results: a = 1 , a = -5

(5.) 6a² + 13a + 6 = 0 here is the problem
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a = [-b ± √b² - 4ac]/(2a) use the quadratic formula
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a = [-13 ± √13² - 4*6*6]/(2*6) make substitutions

a = [-13 ± 5]/12 multiply subtract take sq root

a = -2/3 a = -3/2 add subtract reduce

results: a = -2/3 , a = -3/2

(6.) 2y² = 5y - 3

+ 3 + 3 add 3 to each side
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2y² + 3 = 5y add

-5y -5y subtract 5y from each side
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2y² - 5y + 3 = 0 subtract
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y = [-b ± √b² - 4ac]/(2a) use the quadratic formula
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y = [5 ± √5² - 4*2*3]/(2*2) make substitutions

y = (5 ± 1)/4 multiply subtract and take sq root

y = 3/2 y = 1 add subtract and reduce and divide

results: y = 3/2 , y = 1

(7.) x² + 2x = 1 here is the problem

-1 -1 subtract 1 from each side
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x² + 2x - 1 = 0 subtract
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x = [-b ± √b² - 4ac]/(2a) use the quadratic formula
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x = [-1 ± √2² - 4*1*-1]/(2*1) make substitutions
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x = (-1 ± √8)/2 multiply and add
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x = (-1 ± √4√2)/2 factor like this
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x = (-1 ± 2√2)/2 take the square root of the 4
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x = (-1/2) ± √2 divide thru by 2, cancel
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results: x = (-1/2) + √2 , x = (-1/2) - √2

(8.) x² + 6x + 3 = 0 here is the problem
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x = [-b ± √b² - 4ac]/(2a) use the quadratic formula
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x = [-6 ± √6² - 4*1*3]/(2*1) make substitutions
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x = (-6 ± √12)/2 multiply and subtract
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x = (-6 ± √4√3)/2 factor
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x = (-6 ± 2√3)/2 take sq root of the 4
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x = -3 ± √3 divide thru by 2
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results: x = -3 + √3 , x = -3 - √3

(9.) 2x² + 7x + 2 = 0 here is the problem
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x = [-b ± √b² - 4ac]/(2a) use the quadratic formula
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x = [-7 ± √7² - 4*2*2]/(2*2) make substitutions
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x = (-7 ± √33)/4 multiply subtract and take sq root
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results: x = (-7 + √33)/4 , x = (-7 - √33)/4

(10.) 6x² - 3x - 4 = 0 here is the problem
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x = [-b ± √b² - 4ac]/(2a) use the quadratic formula
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x = [3 ± √3² - 4*6*-4]/(2*6) make substitutions
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x = (3 ± √96)/12 multiply add
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x = (3 ± √16√6)/12 factor
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x = (3 ± 4√6)/12 take sq root of the 16
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x = (1/4) ± (√6/3) separate the fraction and reduce
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results: x = (1/4) + (√6/3) , x = (1/4) - (√6/3)