Solve 16b^2+24b+20=15

$$16 b ^ { 2 } + 24 b + 20 = 15$$

Solve for b

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

$$b=-\frac{1}{4}=-0.25$$

Steps Using Factoring By Grouping

Subtract 15 from both sides.

$$16b^{2}+24b+20-15=0$$

Subtract 15 from 20 to get 5.

$$16b^{2}+24b+5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 16b^{2}+ab+bb+5. To find a and b, set up a system to be solved.

$$a+b=24 ab=16\times 5=80$$

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 80.

$$1,80 2,40 4,20 5,16 8,10$$

Calculate the sum for each pair.

$$1+80=81 2+40=42 4+20=24 5+16=21 8+10=18$$

The solution is the pair that gives sum 24.

$$a=4 b=20$$

Rewrite 16b^{2}+24b+5 as \left(16b^{2}+4b\right)+\left(20b+5\right).

$$\left(16b^{2}+4b\right)+\left(20b+5\right)$$

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

$$4b\left(4b+1\right)+5\left(4b+1\right)$$

Factor out common term 4b+1 by using distributive property.

$$\left(4b+1\right)\left(4b+5\right)$$

To find equation solutions, solve 4b+1=0 and 4b+5=0.

$$b=-\frac{1}{4} b=-\frac{5}{4}$$

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.

$$16b^{2}+24b+20=15$$

Subtract 15 from both sides of the equation.

$$16b^{2}+24b+20-15=15-15$$

Subtracting 15 from itself leaves 0.

$$16b^{2}+24b+20-15=0$$

Subtract 15 from 20.

$$16b^{2}+24b+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, 24 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{-24±\sqrt{24^{2}-4\times 16\times 5}}{2\times 16}$$

Square 24.

$$b=\frac{-24±\sqrt{576-4\times 16\times 5}}{2\times 16}$$

Multiply -4 times 16.

$$b=\frac{-24±\sqrt{576-64\times 5}}{2\times 16}$$

Multiply -64 times 5.

$$b=\frac{-24±\sqrt{576-320}}{2\times 16}$$

Add 576 to -320.

$$b=\frac{-24±\sqrt{256}}{2\times 16}$$

Take the square root of 256.

$$b=\frac{-24±16}{2\times 16}$$

Multiply 2 times 16.

$$b=\frac{-24±16}{32}$$

Now solve the equation b=\frac{-24±16}{32} when ± is plus. Add -24 to 16.

$$b=-\frac{8}{32}$$

Reduce the fraction \frac{-8}{32} to lowest terms by extracting and canceling out 8.

$$b=-\frac{1}{4}$$

Now solve the equation b=\frac{-24±16}{32} when ± is minus. Subtract 16 from -24.

$$b=-\frac{40}{32}$$

Reduce the fraction \frac{-40}{32} to lowest terms by extracting and canceling out 8.

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

The equation is now solved.

$$b=-\frac{1}{4} b=-\frac{5}{4}$$

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.

$$16b^{2}+24b+20=15$$

Subtract 20 from both sides of the equation.

$$16b^{2}+24b+20-20=15-20$$

Subtracting 20 from itself leaves 0.

$$16b^{2}+24b=15-20$$

Subtract 20 from 15.

$$16b^{2}+24b=-5$$

Divide both sides by 16.

$$\frac{16b^{2}+24b}{16}=-\frac{5}{16}$$

Dividing by 16 undoes the multiplication by 16.

$$b^{2}+\frac{24}{16}b=-\frac{5}{16}$$

Reduce the fraction \frac{24}{16} to lowest terms by extracting and canceling out 8.

$$b^{2}+\frac{3}{2}b=-\frac{5}{16}$$

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

$$b^{2}+\frac{3}{2}b+\left(\frac{3}{4}\right)^{2}=-\frac{5}{16}+\left(\frac{3}{4}\right)^{2}$$

Square \frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$b^{2}+\frac{3}{2}b+\frac{9}{16}=\frac{-5+9}{16}$$

Add -\frac{5}{16} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$b^{2}+\frac{3}{2}b+\frac{9}{16}=\frac{1}{4}$$

Factor b^{2}+\frac{3}{2}b+\frac{9}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(b+\frac{3}{4}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(b+\frac{3}{4}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$b+\frac{3}{4}=\frac{1}{2} b+\frac{3}{4}=-\frac{1}{2}$$

Subtract \frac{3}{4} from both sides of the equation.

$$b=-\frac{1}{4} b=-\frac{5}{4}$$