Solve 6p^2+p-2=9p^2-6p-8

$$6 p ^ { 2 } + p - 2 = 9 p ^ { 2 } - 6 p - 8$$

Solve for p

$$p=3$$

$$p=-\frac{2}{3}\approx -0.666666667$$

Steps Using Factoring By Grouping

Subtract 9p^{2} from both sides.

$$6p^{2}+p-2-9p^{2}=-6p-8$$

Combine 6p^{2} and -9p^{2} to get -3p^{2}.

$$-3p^{2}+p-2=-6p-8$$

Add 6p to both sides.

$$-3p^{2}+p-2+6p=-8$$

Combine p and 6p to get 7p.

$$-3p^{2}+7p-2=-8$$

Add 8 to both sides.

$$-3p^{2}+7p-2+8=0$$

Add -2 and 8 to get 6.

$$-3p^{2}+7p+6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3p^{2}+ap+bp+6. To find a and b, set up a system to be solved.

$$a+b=7 ab=-3\times 6=-18$$

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 7.

$$a=9 b=-2$$

Rewrite -3p^{2}+7p+6 as \left(-3p^{2}+9p\right)+\left(-2p+6\right).

$$\left(-3p^{2}+9p\right)+\left(-2p+6\right)$$

Factor out 3p in the first and 2 in the second group.

$$3p\left(-p+3\right)+2\left(-p+3\right)$$

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

$$\left(-p+3\right)\left(3p+2\right)$$

To find equation solutions, solve -p+3=0 and 3p+2=0.

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

Steps Using the Quadratic Formula

Subtract 9p^{2} from both sides.

$$6p^{2}+p-2-9p^{2}=-6p-8$$

Combine 6p^{2} and -9p^{2} to get -3p^{2}.

$$-3p^{2}+p-2=-6p-8$$

Add 6p to both sides.

$$-3p^{2}+p-2+6p=-8$$

Combine p and 6p to get 7p.

$$-3p^{2}+7p-2=-8$$

Add 8 to both sides.

$$-3p^{2}+7p-2+8=0$$

Add -2 and 8 to get 6.

$$-3p^{2}+7p+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 7 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$p=\frac{-7±\sqrt{7^{2}-4\left(-3\right)\times 6}}{2\left(-3\right)}$$

Square 7.

$$p=\frac{-7±\sqrt{49-4\left(-3\right)\times 6}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$p=\frac{-7±\sqrt{49+12\times 6}}{2\left(-3\right)}$$

Multiply 12 times 6.

$$p=\frac{-7±\sqrt{49+72}}{2\left(-3\right)}$$

Add 49 to 72.

$$p=\frac{-7±\sqrt{121}}{2\left(-3\right)}$$

Take the square root of 121.

$$p=\frac{-7±11}{2\left(-3\right)}$$

Multiply 2 times -3.

$$p=\frac{-7±11}{-6}$$

Now solve the equation p=\frac{-7±11}{-6} when ± is plus. Add -7 to 11.

$$p=\frac{4}{-6}$$

Reduce the fraction \frac{4}{-6} to lowest terms by extracting and canceling out 2.

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

Now solve the equation p=\frac{-7±11}{-6} when ± is minus. Subtract 11 from -7.

$$p=-\frac{18}{-6}$$

Divide -18 by -6.

$$p=3$$

The equation is now solved.

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

Steps for Completing the Square

Subtract 9p^{2} from both sides.

$$6p^{2}+p-2-9p^{2}=-6p-8$$

Combine 6p^{2} and -9p^{2} to get -3p^{2}.

$$-3p^{2}+p-2=-6p-8$$

Add 6p to both sides.

$$-3p^{2}+p-2+6p=-8$$

Combine p and 6p to get 7p.

$$-3p^{2}+7p-2=-8$$

Add 2 to both sides.

$$-3p^{2}+7p=-8+2$$

Add -8 and 2 to get -6.

$$-3p^{2}+7p=-6$$

Divide both sides by -3.

$$\frac{-3p^{2}+7p}{-3}=-\frac{6}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$p^{2}+\frac{7}{-3}p=-\frac{6}{-3}$$

Divide 7 by -3.

$$p^{2}-\frac{7}{3}p=-\frac{6}{-3}$$

Divide -6 by -3.

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

Divide -\frac{7}{3}, the coefficient of the x term, by 2 to get -\frac{7}{6}. Then add the square of -\frac{7}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$p^{2}-\frac{7}{3}p+\left(-\frac{7}{6}\right)^{2}=2+\left(-\frac{7}{6}\right)^{2}$$

Square -\frac{7}{6} by squaring both the numerator and the denominator of the fraction.

$$p^{2}-\frac{7}{3}p+\frac{49}{36}=2+\frac{49}{36}$$

Add 2 to \frac{49}{36}.

$$p^{2}-\frac{7}{3}p+\frac{49}{36}=\frac{121}{36}$$

Factor p^{2}-\frac{7}{3}p+\frac{49}{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}.

$$\left(p-\frac{7}{6}\right)^{2}=\frac{121}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(p-\frac{7}{6}\right)^{2}}=\sqrt{\frac{121}{36}}$$

Simplify.

$$p-\frac{7}{6}=\frac{11}{6} p-\frac{7}{6}=-\frac{11}{6}$$

Add \frac{7}{6} to both sides of the equation.

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