Solve for p p^2+p-90=0

$p^{2} + p - 90 = 0$

Factor $p^{2} + p - 90$ 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 $- 90$ and whose sum is $1$ .

$- 9, 10$

Write the factored form using these integers.

$(p - 9) (p + 10) = 0$

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

$p - 9 = 0$

$p + 10 = 0$

Set $p - 9$ equal to $0$ and solve for $p$ .

Set $p - 9$ equal to $0$ .

$p - 9 = 0$

Add $9$ to both sides of the equation.

$p = 9$

Set $p + 10$ equal to $0$ and solve for $p$ .

Set $p + 10$ equal to $0$ .

$p + 10 = 0$

Subtract $10$ from both sides of the equation.

$p = - 10$

The final solution is all the values that make $(p - 9) (p + 10) = 0$ true.

$p = 9, - 10$

Solve for p p^2+p-90=0

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