How do you solve $x^2+12x-64=0$ using the quadratic formula?

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Answer 1 · 2015-07-02T22:10:30

Identify a, b, and c in your equation. Then substitute the values into the quadratic equation. Solve for $x$ .

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$x^2+12x-64=0$

$a=1;$ $b=12;$ $c=-64$

$x=(-12+-sqrt(12^2-(4*1*-64)))/(2*1)$ =

$x=(-12+-sqrt(144+256))/2$ =

$x=(-12+-sqrt(400))/2$ =

$x=(-12+-20)/2$

$x=(-12+20)/2$ =

$x=8/2$ =

$x=4$

$x=(-12-20)/2$ =

$x=-32/2$ =

$x=-16$

$x=4, -16$

How do you solve x^2+12x-64=0x^2+12x-64=0 using the quadratic formula?