Solve for x x^2=x+20

$x^{2} = x + 20$

Step 1

Subtract $x$ from both sides of the equation.

$x^{2} - x = 20$

Step 2

Subtract $20$ from both sides of the equation.

$x^{2} - x - 20 = 0$

Step 3

Factor $x^{2} - x - 20$ 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 $- 20$ and whose sum is $- 1$ .

$- 5, 4$

Write the factored form using these integers.

$(x - 5) (x + 4) = 0$

Step 4

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

$x - 5 = 0$

$x + 4 = 0$

Step 5

Set $x - 5$ equal to $0$ and solve for $x$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Step 6

Set $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 7

The final solution is all the values that make $(x - 5) (x + 4) = 0$ true.

$x = 5, - 4$

Solve for x x^2=x+20

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