How do you solve $x^2 - 4x - 32 = 0$ ?

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Answer 1 · 2018-03-23T23:22:56

The root of the equation is $x = 2+-6$ or $x = 8$ and $x = -4$ .

For an equation in the form of $ax^2 + bx + c= 0$ , the quadratic formula ( $x=(-b+-sqrt(b^2-4ac))/(2a)$ ) can be applied.

Plugging in $1$ for $a$ , $-4$ for $b$ , and $-32$ for $c$ :

$x=(-(-4)+-sqrt((-4)^2-4(1)(-32)))/(2(1))$

$x=(4+-sqrt(16-4(-32)))/2$

$x=(4+-sqrt(16-(-128)))/2$

$x=(4+-sqrt(144))/2$

$x=4/2+-sqrt(144)/2$

$x=2+-12/2$

$x=2 +- 6$

Doing addition:

$x=2+6 = 8$

Doing subtraction:

$x=2-6 = -4$

How do you solve x^2 - 4x - 32 = 0x^2 - 4x - 32 = 0?