Solve -3p^2+10p+5=-8p^2

$$- 3 p ^ { 2 } + 10 p + 5 = - 8 p ^ { 2 }$$

Solve for p

$$p=-1$$

Steps Using Factoring By Grouping

Add 8p^{2} to both sides.

$$-3p^{2}+10p+5+8p^{2}=0$$

Combine -3p^{2} and 8p^{2} to get 5p^{2}.

$$5p^{2}+10p+5=0$$

Divide both sides by 5.

$$p^{2}+2p+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as p^{2}+ap+bp+1. To find a and b, set up a system to be solved.

$$a+b=2 ab=1\times 1=1$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=1$$

Rewrite p^{2}+2p+1 as \left(p^{2}+p\right)+\left(p+1\right).

$$\left(p^{2}+p\right)+\left(p+1\right)$$

Factor out p in p^{2}+p.

$$p\left(p+1\right)+p+1$$

Factor out common term p+1 by using distributive property.

$$\left(p+1\right)\left(p+1\right)$$

Rewrite as a binomial square.

$$\left(p+1\right)^{2}$$

To find equation solution, solve p+1=0.

$$p=-1$$

Steps Using the Quadratic Formula

Add 8p^{2} to both sides.

$$-3p^{2}+10p+5+8p^{2}=0$$

Combine -3p^{2} and 8p^{2} to get 5p^{2}.

$$5p^{2}+10p+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 10 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$p=\frac{-10±\sqrt{10^{2}-4\times 5\times 5}}{2\times 5}$$

Square 10.

$$p=\frac{-10±\sqrt{100-4\times 5\times 5}}{2\times 5}$$

Multiply -4 times 5.

$$p=\frac{-10±\sqrt{100-20\times 5}}{2\times 5}$$

Multiply -20 times 5.

$$p=\frac{-10±\sqrt{100-100}}{2\times 5}$$

Add 100 to -100.

$$p=\frac{-10±\sqrt{0}}{2\times 5}$$

Take the square root of 0.

$$p=-\frac{10}{2\times 5}$$

Multiply 2 times 5.

$$p=-\frac{10}{10}$$

Divide -10 by 10.

$$p=-1$$

Steps for Completing the Square

Add 8p^{2} to both sides.

$$-3p^{2}+10p+5+8p^{2}=0$$

Combine -3p^{2} and 8p^{2} to get 5p^{2}.

$$5p^{2}+10p+5=0$$

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

$$5p^{2}+10p=-5$$

Divide both sides by 5.

$$\frac{5p^{2}+10p}{5}=-\frac{5}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$p^{2}+\frac{10}{5}p=-\frac{5}{5}$$

Divide 10 by 5.

$$p^{2}+2p=-\frac{5}{5}$$

Divide -5 by 5.

$$p^{2}+2p=-1$$

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$p^{2}+2p+1^{2}=-1+1^{2}$$

Square 1.

$$p^{2}+2p+1=-1+1$$

Add -1 to 1.

$$p^{2}+2p+1=0$$

Factor p^{2}+2p+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(p+1\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(p+1\right)^{2}}=\sqrt{0}$$

Simplify.

$$p+1=0 p+1=0$$

Subtract 1 from both sides of the equation.

$$p=-1 p=-1$$

The equation is now solved. Solutions are the same.

$$p=-1$$