Solve (8y+6)^2=100

$$( 8 y + 6 ) ^ { 2 } = 100$$

Solve for y

$$y=\frac{1}{2}=0.5$$

$$y=-2$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(8y+6\right)^{2}.

$$64y^{2}+96y+36=100$$

Subtract 100 from both sides.

$$64y^{2}+96y+36-100=0$$

Subtract 100 from 36 to get -64.

$$64y^{2}+96y-64=0$$

Divide both sides by 32.

$$2y^{2}+3y-2=0$$

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

$$a+b=3 ab=2\left(-2\right)=-4$$

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 -4.

$$-1,4 -2,2$$

Calculate the sum for each pair.

$$-1+4=3 -2+2=0$$

The solution is the pair that gives sum 3.

$$a=-1 b=4$$

Rewrite 2y^{2}+3y-2 as \left(2y^{2}-y\right)+\left(4y-2\right).

$$\left(2y^{2}-y\right)+\left(4y-2\right)$$

Factor out y in the first and 2 in the second group.

$$y\left(2y-1\right)+2\left(2y-1\right)$$

Factor out common term 2y-1 by using distributive property.

$$\left(2y-1\right)\left(y+2\right)$$

To find equation solutions, solve 2y-1=0 and y+2=0.

$$y=\frac{1}{2} y=-2$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(8y+6\right)^{2}.

$$64y^{2}+96y+36=100$$

Subtract 100 from both sides.

$$64y^{2}+96y+36-100=0$$

Subtract 100 from 36 to get -64.

$$64y^{2}+96y-64=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 64 for a, 96 for b, and -64 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-96±\sqrt{96^{2}-4\times 64\left(-64\right)}}{2\times 64}$$

Square 96.

$$y=\frac{-96±\sqrt{9216-4\times 64\left(-64\right)}}{2\times 64}$$

Multiply -4 times 64.

$$y=\frac{-96±\sqrt{9216-256\left(-64\right)}}{2\times 64}$$

Multiply -256 times -64.

$$y=\frac{-96±\sqrt{9216+16384}}{2\times 64}$$

Add 9216 to 16384.

$$y=\frac{-96±\sqrt{25600}}{2\times 64}$$

Take the square root of 25600.

$$y=\frac{-96±160}{2\times 64}$$

Multiply 2 times 64.

$$y=\frac{-96±160}{128}$$

Now solve the equation y=\frac{-96±160}{128} when ± is plus. Add -96 to 160.

$$y=\frac{64}{128}$$

Reduce the fraction \frac{64}{128} to lowest terms by extracting and canceling out 64.

$$y=\frac{1}{2}$$

Now solve the equation y=\frac{-96±160}{128} when ± is minus. Subtract 160 from -96.

$$y=-\frac{256}{128}$$

Divide -256 by 128.

$$y=-2$$

The equation is now solved.

$$y=\frac{1}{2} y=-2$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(8y+6\right)^{2}.

$$64y^{2}+96y+36=100$$

Subtract 36 from both sides.

$$64y^{2}+96y=100-36$$

Subtract 36 from 100 to get 64.

$$64y^{2}+96y=64$$

Divide both sides by 64.

$$\frac{64y^{2}+96y}{64}=\frac{64}{64}$$

Dividing by 64 undoes the multiplication by 64.

$$y^{2}+\frac{96}{64}y=\frac{64}{64}$$

Reduce the fraction \frac{96}{64} to lowest terms by extracting and canceling out 32.

$$y^{2}+\frac{3}{2}y=\frac{64}{64}$$

Divide 64 by 64.

$$y^{2}+\frac{3}{2}y=1$$

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.

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

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

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

Add 1 to \frac{9}{16}.

$$y^{2}+\frac{3}{2}y+\frac{9}{16}=\frac{25}{16}$$

Factor y^{2}+\frac{3}{2}y+\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}.

$$\left(y+\frac{3}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+\frac{3}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$y+\frac{3}{4}=\frac{5}{4} y+\frac{3}{4}=-\frac{5}{4}$$

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

$$y=\frac{1}{2} y=-2$$