Solve 200+115p=-15p^2

$$200 + 115 p = - 15 p ^ { 2 }$$

Solve for p

$$p = -\frac{8}{3} = -2\frac{2}{3} \approx -2.666666667$$

$$p=-5$$

Steps Using the Quadratic Formula

Add 15p^{2} to both sides.

$$200+115p+15p^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$15p^{2}+115p+200=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 15 for a, 115 for b, and 200 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$p=\frac{-115±\sqrt{115^{2}-4\times 15\times 200}}{2\times 15}$$

Square 115.

$$p=\frac{-115±\sqrt{13225-4\times 15\times 200}}{2\times 15}$$

Multiply -4 times 15.

$$p=\frac{-115±\sqrt{13225-60\times 200}}{2\times 15}$$

Multiply -60 times 200.

$$p=\frac{-115±\sqrt{13225-12000}}{2\times 15}$$

Add 13225 to -12000.

$$p=\frac{-115±\sqrt{1225}}{2\times 15}$$

Take the square root of 1225.

$$p=\frac{-115±35}{2\times 15}$$

Multiply 2 times 15.

$$p=\frac{-115±35}{30}$$

Now solve the equation p=\frac{-115±35}{30} when ± is plus. Add -115 to 35.

$$p=-\frac{80}{30}$$

Reduce the fraction \frac{-80}{30} to lowest terms by extracting and canceling out 10.

$$p=-\frac{8}{3}$$

Now solve the equation p=\frac{-115±35}{30} when ± is minus. Subtract 35 from -115.

$$p=-\frac{150}{30}$$

Divide -150 by 30.

$$p=-5$$

The equation is now solved.

$$p=-\frac{8}{3} p=-5$$

Steps for Completing the Square

Add 15p^{2} to both sides.

$$200+115p+15p^{2}=0$$

Subtract 200 from both sides. Anything subtracted from zero gives its negation.

$$115p+15p^{2}=-200$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$15p^{2}+115p=-200$$

Divide both sides by 15.

$$\frac{15p^{2}+115p}{15}=-\frac{200}{15}$$

Dividing by 15 undoes the multiplication by 15.

$$p^{2}+\frac{115}{15}p=-\frac{200}{15}$$

Reduce the fraction \frac{115}{15} to lowest terms by extracting and canceling out 5.

$$p^{2}+\frac{23}{3}p=-\frac{200}{15}$$

Reduce the fraction \frac{-200}{15} to lowest terms by extracting and canceling out 5.

$$p^{2}+\frac{23}{3}p=-\frac{40}{3}$$

Divide \frac{23}{3}, the coefficient of the x term, by 2 to get \frac{23}{6}. Then add the square of \frac{23}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$p^{2}+\frac{23}{3}p+\left(\frac{23}{6}\right)^{2}=-\frac{40}{3}+\left(\frac{23}{6}\right)^{2}$$

Square \frac{23}{6} by squaring both the numerator and the denominator of the fraction.

$$p^{2}+\frac{23}{3}p+\frac{529}{36}=-\frac{40}{3}+\frac{529}{36}$$

Add -\frac{40}{3} to \frac{529}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$p^{2}+\frac{23}{3}p+\frac{529}{36}=\frac{49}{36}$$

Factor p^{2}+\frac{23}{3}p+\frac{529}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(p+\frac{23}{6}\right)^{2}=\frac{49}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(p+\frac{23}{6}\right)^{2}}=\sqrt{\frac{49}{36}}$$

Simplify.

$$p+\frac{23}{6}=\frac{7}{6} p+\frac{23}{6}=-\frac{7}{6}$$

Subtract \frac{23}{6} from both sides of the equation.

$$p=-\frac{8}{3} p=-5$$