# How do you solve x^2+4x=21?

Apr 8, 2017

$x = - 7$ or $x = 3$

#### Explanation:

let ${x}^{2} + 4 x$ = 21
Moving the 21 to the other side we get ${x}^{2} + 4 x - 21 = 0$
Now, we want two numbers that multiply to give -21 and and to give 4

These two numbers are 7 and -3.
7 x -3 = 21
7+-3 = 4

Factorizing this we get:

$\left(x + 7\right) \left(x - 3\right)$

Now we make both brackets equal to zero
$x + 7$ = 0 or $x - 3$=0
$x$ = -7 $x$=3

Apr 8, 2017

See below.

#### Explanation:

We move everything onto one side, and factor.

${x}^{2} + 4 x - 21 = 0$

${x}^{2} - 3 x + 7 x - 21 = 0$

$x \left(x - 3\right) + 7 \left(x - 3\right) = 0$

$\left(x + 7\right) \left(x - 3\right) = 0$

$x = - 7 , 3$