Solve p^2=-3p+18

$$p ^ { 2 } = - 3 p + 18$$

Solve for p

$$p=-6$$

$$p=3$$

Steps Using Factoring

Add 3p to both sides.

$$p^{2}+3p=18$$

Subtract 18 from both sides.

$$p^{2}+3p-18=0$$

To solve the equation, factor p^{2}+3p-18 using formula p^{2}+\left(a+b\right)p+ab=\left(p+a\right)\left(p+b\right). To find a and b, set up a system to be solved.

$$a+b=3 ab=-18$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -18.

$$-1,18 -2,9 -3,6$$

Calculate the sum for each pair.

$$-1+18=17 -2+9=7 -3+6=3$$

The solution is the pair that gives sum 3.

$$a=-3 b=6$$

Rewrite factored expression \left(p+a\right)\left(p+b\right) using the obtained values.

$$\left(p-3\right)\left(p+6\right)$$

To find equation solutions, solve p-3=0 and p+6=0.

$$p=3 p=-6$$

Steps Using Factoring By Grouping

Add 3p to both sides.

$$p^{2}+3p=18$$

Subtract 18 from both sides.

$$p^{2}+3p-18=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as p^{2}+ap+bp-18. To find a and b, set up a system to be solved.

$$a+b=3 ab=1\left(-18\right)=-18$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -18.

$$-1,18 -2,9 -3,6$$

Calculate the sum for each pair.

$$-1+18=17 -2+9=7 -3+6=3$$

The solution is the pair that gives sum 3.

$$a=-3 b=6$$

Rewrite p^{2}+3p-18 as \left(p^{2}-3p\right)+\left(6p-18\right).

$$\left(p^{2}-3p\right)+\left(6p-18\right)$$

Factor out p in the first and 6 in the second group.

$$p\left(p-3\right)+6\left(p-3\right)$$

Factor out common term p-3 by using distributive property.

$$\left(p-3\right)\left(p+6\right)$$

To find equation solutions, solve p-3=0 and p+6=0.

$$p=3 p=-6$$

Steps Using the Quadratic Formula

Add 3p to both sides.

$$p^{2}+3p=18$$

Subtract 18 from both sides.

$$p^{2}+3p-18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 3 for b, and -18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$p=\frac{-3±\sqrt{3^{2}-4\left(-18\right)}}{2}$$

Square 3.

$$p=\frac{-3±\sqrt{9-4\left(-18\right)}}{2}$$

Multiply -4 times -18.

$$p=\frac{-3±\sqrt{9+72}}{2}$$

Add 9 to 72.

$$p=\frac{-3±\sqrt{81}}{2}$$

Take the square root of 81.

$$p=\frac{-3±9}{2}$$

Now solve the equation p=\frac{-3±9}{2} when ± is plus. Add -3 to 9.

$$p=\frac{6}{2}$$

Divide 6 by 2.

$$p=3$$

Now solve the equation p=\frac{-3±9}{2} when ± is minus. Subtract 9 from -3.

$$p=-\frac{12}{2}$$

Divide -12 by 2.

$$p=-6$$

The equation is now solved.

$$p=3 p=-6$$

Steps for Completing the Square

Add 3p to both sides.

$$p^{2}+3p=18$$

Divide 3, the coefficient of the x term, by 2 to get \frac{3}{2}. Then add the square of \frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$p^{2}+3p+\left(\frac{3}{2}\right)^{2}=18+\left(\frac{3}{2}\right)^{2}$$

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$p^{2}+3p+\frac{9}{4}=18+\frac{9}{4}$$

Add 18 to \frac{9}{4}.

$$p^{2}+3p+\frac{9}{4}=\frac{81}{4}$$

Factor p^{2}+3p+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(p+\frac{3}{2}\right)^{2}=\frac{81}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(p+\frac{3}{2}\right)^{2}}=\sqrt{\frac{81}{4}}$$

Simplify.

$$p+\frac{3}{2}=\frac{9}{2} p+\frac{3}{2}=-\frac{9}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$p=3 p=-6$$