Solve for x x^2+4x=21

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

Step 1

Subtract $21$ from both sides of the equation.

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

Step 2

Factor $x^{2} + 4 x - 21$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 21$ and whose sum is $4$ .

$- 3, 7$

Write the factored form using these integers.

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

Step 3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 3 = 0$

$x + 7 = 0$

Step 4

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Step 5

Set $x + 7$ equal to $0$ and solve for $x$ .

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Subtract $7$ from both sides of the equation.

$x = - 7$

Step 6

The final solution is all the values that make $(x - 3) (x + 7) = 0$ true.

$x = 3, - 7$

Solve for x x^2+4x=21

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