Solve 3p^2+7=-4p

$$3 p ^ { 2 } + 7 = - 4 p$$

Solve for p

$$p=\frac{-2+\sqrt{17}i}{3}\approx -0.666666667+1.374368542i$$

$$p=\frac{-\sqrt{17}i-2}{3}\approx -0.666666667-1.374368542i$$

Steps Using the Quadratic Formula

Add 4p to both sides.

$$3p^{2}+7+4p=0$$

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.

$$3p^{2}+4p+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 4 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$p=\frac{-4±\sqrt{4^{2}-4\times 3\times 7}}{2\times 3}$$

Square 4.

$$p=\frac{-4±\sqrt{16-4\times 3\times 7}}{2\times 3}$$

Multiply -4 times 3.

$$p=\frac{-4±\sqrt{16-12\times 7}}{2\times 3}$$

Multiply -12 times 7.

$$p=\frac{-4±\sqrt{16-84}}{2\times 3}$$

Add 16 to -84.

$$p=\frac{-4±\sqrt{-68}}{2\times 3}$$

Take the square root of -68.

$$p=\frac{-4±2\sqrt{17}i}{2\times 3}$$

Multiply 2 times 3.

$$p=\frac{-4±2\sqrt{17}i}{6}$$

Now solve the equation p=\frac{-4±2\sqrt{17}i}{6} when ± is plus. Add -4 to 2i\sqrt{17}.

$$p=\frac{-4+2\sqrt{17}i}{6}$$

Divide -4+2i\sqrt{17} by 6.

$$p=\frac{-2+\sqrt{17}i}{3}$$

Now solve the equation p=\frac{-4±2\sqrt{17}i}{6} when ± is minus. Subtract 2i\sqrt{17} from -4.

$$p=\frac{-2\sqrt{17}i-4}{6}$$

Divide -4-2i\sqrt{17} by 6.

$$p=\frac{-\sqrt{17}i-2}{3}$$

The equation is now solved.

$$p=\frac{-2+\sqrt{17}i}{3} p=\frac{-\sqrt{17}i-2}{3}$$

Steps for Completing the Square

Add 4p to both sides.

$$3p^{2}+7+4p=0$$

Subtract 7 from both sides. Anything subtracted from zero gives its negation.

$$3p^{2}+4p=-7$$

Divide both sides by 3.

$$\frac{3p^{2}+4p}{3}=-\frac{7}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$p^{2}+\frac{4}{3}p=-\frac{7}{3}$$

Divide \frac{4}{3}, the coefficient of the x term, by 2 to get \frac{2}{3}. Then add the square of \frac{2}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$p^{2}+\frac{4}{3}p+\left(\frac{2}{3}\right)^{2}=-\frac{7}{3}+\left(\frac{2}{3}\right)^{2}$$

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

$$p^{2}+\frac{4}{3}p+\frac{4}{9}=-\frac{7}{3}+\frac{4}{9}$$

Add -\frac{7}{3} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$p^{2}+\frac{4}{3}p+\frac{4}{9}=-\frac{17}{9}$$

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

$$\left(p+\frac{2}{3}\right)^{2}=-\frac{17}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(p+\frac{2}{3}\right)^{2}}=\sqrt{-\frac{17}{9}}$$

Simplify.

$$p+\frac{2}{3}=\frac{\sqrt{17}i}{3} p+\frac{2}{3}=-\frac{\sqrt{17}i}{3}$$

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

$$p=\frac{-2+\sqrt{17}i}{3} p=\frac{-\sqrt{17}i-2}{3}$$