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

Question

1327 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-08T21:12:40

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

let $x^2 + 4x$ = 21
Moving the 21 to the other side we get $x^2 +4x -21 = 0$
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:

$(x+7)(x-3)$

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

Answer 2 · 2017-04-08T21:13:01

See below.

We move everything onto one side, and factor.

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

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

$x(x-3)+7(x-3)=0$

$(x+7)(x-3)=0$

$x=-7,3$

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