Solve for x x^2=-1x+42

$x^{2} = - 1 x + 42$

Rewrite $- 1 x$ as $- x$ .

$x^{2} = - x + 42$

Add $x$ to both sides of the equation.

$x^{2} + x = 42$

Subtract $42$ from both sides of the equation.

$x^{2} + x - 42 = 0$

Factor $x^{2} + x - 42$ 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 $- 42$ and whose sum is $1$ .

$- 6, 7$

Write the factored form using these integers.

$(x - 6) (x + 7) = 0$

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

$x - 6 = 0$

$x + 7 = 0$

Set $x - 6$ equal to $0$ and solve for $x$ .

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

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.

$x = - 7$

The final solution is all the values that make $(x - 6) (x + 7) = 0$ true.

$x = 6, - 7$

Solve for x x^2=-1x+42

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