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

Question

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

1 Answer

Impact of this question

8812 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-23T23:22:56

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

Explanation:

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

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

$x = \frac{- (- 4) \pm \sqrt{{(- 4)}^{2} - 4 (1) (- 32)}}{2 (1)}$ $x=(-(-4)+-sqrt((-4)^2-4(1)(-32)))/(2(1))$

$x = \frac{4 \pm \sqrt{16 - 4 (- 32)}}{2}$ $x=(4+-sqrt(16-4(-32)))/2$

$x = \frac{4 \pm \sqrt{16 - (- 128)}}{2}$ $x=(4+-sqrt(16-(-128)))/2$

$x = \frac{4 \pm \sqrt{144}}{2}$ $x=(4+-sqrt(144))/2$

$x = \frac{4}{2} \pm \frac{\sqrt{144}}{2}$ $x=4/2+-sqrt(144)/2$

$x = 2 \pm \frac{12}{2}$ $x=2+-12/2$

$x = 2 \pm 6$ $x=2 +- 6$

Doing addition:

$x = 2 + 6 = 8$ $x=2+6 = 8$

Doing subtraction:

$x = 2 - 6 = - 4$ $x=2-6 = -4$

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve x2−4x−32=0x^2 - 4x - 32 = 0?

1 Answer

Explanation:

Related questions

Impact of this question

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