Solve (2y-3)^2-64=0

$$( 2 y - 3 ) ^ { 2 } - 64 = 0$$

Solve for y

$$y = \frac{11}{2} = 5\frac{1}{2} = 5.5$$

$$y = -\frac{5}{2} = -2\frac{1}{2} = -2.5$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2y-3\right)^{2}.

$$4y^{2}-12y+9-64=0$$

Subtract 64 from 9 to get -55.

$$4y^{2}-12y-55=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4y^{2}+ay+by-55. To find a and b, set up a system to be solved.

$$a+b=-12 ab=4\left(-55\right)=-220$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -220.

$$1,-220 2,-110 4,-55 5,-44 10,-22 11,-20$$

Calculate the sum for each pair.

$$1-220=-219 2-110=-108 4-55=-51 5-44=-39 10-22=-12 11-20=-9$$

The solution is the pair that gives sum -12.

$$a=-22 b=10$$

Rewrite 4y^{2}-12y-55 as \left(4y^{2}-22y\right)+\left(10y-55\right).

$$\left(4y^{2}-22y\right)+\left(10y-55\right)$$

Factor out 2y in the first and 5 in the second group.

$$2y\left(2y-11\right)+5\left(2y-11\right)$$

Factor out common term 2y-11 by using distributive property.

$$\left(2y-11\right)\left(2y+5\right)$$

To find equation solutions, solve 2y-11=0 and 2y+5=0.

$$y=\frac{11}{2} y=-\frac{5}{2}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2y-3\right)^{2}.

$$4y^{2}-12y+9-64=0$$

Subtract 64 from 9 to get -55.

$$4y^{2}-12y-55=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -12 for b, and -55 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 4\left(-55\right)}}{2\times 4}$$

Square -12.

$$y=\frac{-\left(-12\right)±\sqrt{144-4\times 4\left(-55\right)}}{2\times 4}$$

Multiply -4 times 4.

$$y=\frac{-\left(-12\right)±\sqrt{144-16\left(-55\right)}}{2\times 4}$$

Multiply -16 times -55.

$$y=\frac{-\left(-12\right)±\sqrt{144+880}}{2\times 4}$$

Add 144 to 880.

$$y=\frac{-\left(-12\right)±\sqrt{1024}}{2\times 4}$$

Take the square root of 1024.

$$y=\frac{-\left(-12\right)±32}{2\times 4}$$

The opposite of -12 is 12.

$$y=\frac{12±32}{2\times 4}$$

Multiply 2 times 4.

$$y=\frac{12±32}{8}$$

Now solve the equation y=\frac{12±32}{8} when ± is plus. Add 12 to 32.

$$y=\frac{44}{8}$$

Reduce the fraction \frac{44}{8} to lowest terms by extracting and canceling out 4.

$$y=\frac{11}{2}$$

Now solve the equation y=\frac{12±32}{8} when ± is minus. Subtract 32 from 12.

$$y=-\frac{20}{8}$$

Reduce the fraction \frac{-20}{8} to lowest terms by extracting and canceling out 4.

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

The equation is now solved.

$$y=\frac{11}{2} y=-\frac{5}{2}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2y-3\right)^{2}.

$$4y^{2}-12y+9-64=0$$

Subtract 64 from 9 to get -55.

$$4y^{2}-12y-55=0$$

Add 55 to both sides. Anything plus zero gives itself.

$$4y^{2}-12y=55$$

Divide both sides by 4.

$$\frac{4y^{2}-12y}{4}=\frac{55}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$y^{2}+\left(-\frac{12}{4}\right)y=\frac{55}{4}$$

Divide -12 by 4.

$$y^{2}-3y=\frac{55}{4}$$

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

$$y^{2}-3y+\left(-\frac{3}{2}\right)^{2}=\frac{55}{4}+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-3y+\frac{9}{4}=\frac{55+9}{4}$$

Add \frac{55}{4} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}-3y+\frac{9}{4}=16$$

Factor y^{2}-3y+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-\frac{3}{2}\right)^{2}=16$$

Take the square root of both sides of the equation.

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

Simplify.

$$y-\frac{3}{2}=4 y-\frac{3}{2}=-4$$

Add \frac{3}{2} to both sides of the equation.

$$y=\frac{11}{2} y=-\frac{5}{2}$$