Solve 20p^2+33p+16=6

$$20 p ^ { 2 } + 33 p + 16 = 6$$

Solve for p

$$p = -\frac{5}{4} = -1\frac{1}{4} = -1.25$$

$$p=-\frac{2}{5}=-0.4$$

Steps Using Factoring By Grouping

Subtract 6 from both sides.

$$20p^{2}+33p+16-6=0$$

Subtract 6 from 16 to get 10.

$$20p^{2}+33p+10=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 20p^{2}+ap+bp+10. To find a and b, set up a system to be solved.

$$a+b=33 ab=20\times 10=200$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 200.

$$1,200 2,100 4,50 5,40 8,25 10,20$$

Calculate the sum for each pair.

$$1+200=201 2+100=102 4+50=54 5+40=45 8+25=33 10+20=30$$

The solution is the pair that gives sum 33.

$$a=8 b=25$$

Rewrite 20p^{2}+33p+10 as \left(20p^{2}+8p\right)+\left(25p+10\right).

$$\left(20p^{2}+8p\right)+\left(25p+10\right)$$

Factor out 4p in the first and 5 in the second group.

$$4p\left(5p+2\right)+5\left(5p+2\right)$$

Factor out common term 5p+2 by using distributive property.

$$\left(5p+2\right)\left(4p+5\right)$$

To find equation solutions, solve 5p+2=0 and 4p+5=0.

$$p=-\frac{2}{5} p=-\frac{5}{4}$$

Steps Using the Quadratic Formula

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.

$$20p^{2}+33p+16=6$$

Subtract 6 from both sides of the equation.

$$20p^{2}+33p+16-6=6-6$$

Subtracting 6 from itself leaves 0.

$$20p^{2}+33p+16-6=0$$

Subtract 6 from 16.

$$20p^{2}+33p+10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 20 for a, 33 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$p=\frac{-33±\sqrt{33^{2}-4\times 20\times 10}}{2\times 20}$$

Square 33.

$$p=\frac{-33±\sqrt{1089-4\times 20\times 10}}{2\times 20}$$

Multiply -4 times 20.

$$p=\frac{-33±\sqrt{1089-80\times 10}}{2\times 20}$$

Multiply -80 times 10.

$$p=\frac{-33±\sqrt{1089-800}}{2\times 20}$$

Add 1089 to -800.

$$p=\frac{-33±\sqrt{289}}{2\times 20}$$

Take the square root of 289.

$$p=\frac{-33±17}{2\times 20}$$

Multiply 2 times 20.

$$p=\frac{-33±17}{40}$$

Now solve the equation p=\frac{-33±17}{40} when ± is plus. Add -33 to 17.

$$p=-\frac{16}{40}$$

Reduce the fraction \frac{-16}{40} to lowest terms by extracting and canceling out 8.

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

Now solve the equation p=\frac{-33±17}{40} when ± is minus. Subtract 17 from -33.

$$p=-\frac{50}{40}$$

Reduce the fraction \frac{-50}{40} to lowest terms by extracting and canceling out 10.

$$p=-\frac{5}{4}$$

The equation is now solved.

$$p=-\frac{2}{5} p=-\frac{5}{4}$$

Steps for Completing the Square

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.

$$20p^{2}+33p+16=6$$

Subtract 16 from both sides of the equation.

$$20p^{2}+33p+16-16=6-16$$

Subtracting 16 from itself leaves 0.

$$20p^{2}+33p=6-16$$

Subtract 16 from 6.

$$20p^{2}+33p=-10$$

Divide both sides by 20.

$$\frac{20p^{2}+33p}{20}=-\frac{10}{20}$$

Dividing by 20 undoes the multiplication by 20.

$$p^{2}+\frac{33}{20}p=-\frac{10}{20}$$

Reduce the fraction \frac{-10}{20} to lowest terms by extracting and canceling out 10.

$$p^{2}+\frac{33}{20}p=-\frac{1}{2}$$

Divide \frac{33}{20}, the coefficient of the x term, by 2 to get \frac{33}{40}. Then add the square of \frac{33}{40} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$p^{2}+\frac{33}{20}p+\left(\frac{33}{40}\right)^{2}=-\frac{1}{2}+\left(\frac{33}{40}\right)^{2}$$

Square \frac{33}{40} by squaring both the numerator and the denominator of the fraction.

$$p^{2}+\frac{33}{20}p+\frac{1089}{1600}=-\frac{1}{2}+\frac{1089}{1600}$$

Add -\frac{1}{2} to \frac{1089}{1600} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$p^{2}+\frac{33}{20}p+\frac{1089}{1600}=\frac{289}{1600}$$

Factor p^{2}+\frac{33}{20}p+\frac{1089}{1600}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(p+\frac{33}{40}\right)^{2}=\frac{289}{1600}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(p+\frac{33}{40}\right)^{2}}=\sqrt{\frac{289}{1600}}$$

Simplify.

$$p+\frac{33}{40}=\frac{17}{40} p+\frac{33}{40}=-\frac{17}{40}$$

Subtract \frac{33}{40} from both sides of the equation.

$$p=-\frac{2}{5} p=-\frac{5}{4}$$