How do you solve the quadratic $(x - 5) (x + 8) = - 20$ $(x-5)(x+8)=-20$ using any method?

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How do you solve the quadratic $(x - 5) (x + 8) = - 20$ $(x-5)(x+8)=-20$ using any method?

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Answer 1 · 2016-08-16T16:38:02

$x = \frac{1}{2} (- 3 \pm \sqrt{89})$ $x = 1/2(-3+-sqrt(89))$

Explanation:

$- 20 = (x - 5) (x + 8) = x^{2} + 3 x - 40$ $-20 = (x-5)(x+8)=x^2+3x-40$

Add $20$ $20$ to both ends to get:

$0 = x^{2} + 3 x - 20$ $0 = x^2+3x-20$

Multiply by $4$ $4$ and complete the square:

$0 = 4 x^{2} + 12 x - 80$ $0 = 4x^2+12x-80$

$= {(2 x)}^{2} + 2 (3) (2 x) + 3^{2} - 89$ $=(2x)^2+2(3)(2x)+3^2-89$

$= {(2 x + 3)}^{2} - {(\sqrt{89})}^{2}$ $=(2x+3)^2-(sqrt(89))^2$

$= (2 x + 3 - \sqrt{89}) (2 x + 3 + \sqrt{89})$ $=(2x+3-sqrt(89))(2x+3+sqrt(89))$

Hence $x = \frac{1}{2} (- 3 \pm \sqrt{89})$ $x = 1/2(-3+-sqrt(89))$

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How do you solve the quadratic $(x - 5) (x + 8) = - 20$ $(x-5)(x+8)=-20$ using any method?

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How do you solve the quadratic $(x - 5) (x + 8) = - 20$ $(x-5)(x+8)=-20$ using any method?