Solve (2x+5)^2=81

$$( 2 x + 5 ) ^ { 2 } = 81$$

Solve for x

$$x=2$$

$$x=-7$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+5\right)^{2}.

$$4x^{2}+20x+25=81$$

Subtract 81 from both sides.

$$4x^{2}+20x+25-81=0$$

Subtract 81 from 25 to get -56.

$$4x^{2}+20x-56=0$$

Divide both sides by 4.

$$x^{2}+5x-14=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-14. To find a and b, set up a system to be solved.

$$a+b=5 ab=1\left(-14\right)=-14$$

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 -14.

$$-1,14 -2,7$$

Calculate the sum for each pair.

$$-1+14=13 -2+7=5$$

The solution is the pair that gives sum 5.

$$a=-2 b=7$$

Rewrite x^{2}+5x-14 as \left(x^{2}-2x\right)+\left(7x-14\right).

$$\left(x^{2}-2x\right)+\left(7x-14\right)$$

Factor out x in the first and 7 in the second group.

$$x\left(x-2\right)+7\left(x-2\right)$$

Factor out common term x-2 by using distributive property.

$$\left(x-2\right)\left(x+7\right)$$

To find equation solutions, solve x-2=0 and x+7=0.

$$x=2 x=-7$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+5\right)^{2}.

$$4x^{2}+20x+25=81$$

Subtract 81 from both sides.

$$4x^{2}+20x+25-81=0$$

Subtract 81 from 25 to get -56.

$$4x^{2}+20x-56=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 20 for b, and -56 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-20±\sqrt{20^{2}-4\times 4\left(-56\right)}}{2\times 4}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\times 4\left(-56\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-20±\sqrt{400-16\left(-56\right)}}{2\times 4}$$

Multiply -16 times -56.

$$x=\frac{-20±\sqrt{400+896}}{2\times 4}$$

Add 400 to 896.

$$x=\frac{-20±\sqrt{1296}}{2\times 4}$$

Take the square root of 1296.

$$x=\frac{-20±36}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-20±36}{8}$$

Now solve the equation x=\frac{-20±36}{8} when ± is plus. Add -20 to 36.

$$x=\frac{16}{8}$$

Divide 16 by 8.

$$x=2$$

Now solve the equation x=\frac{-20±36}{8} when ± is minus. Subtract 36 from -20.

$$x=-\frac{56}{8}$$

Divide -56 by 8.

$$x=-7$$

The equation is now solved.

$$x=2 x=-7$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+5\right)^{2}.

$$4x^{2}+20x+25=81$$

Subtract 25 from both sides.

$$4x^{2}+20x=81-25$$

Subtract 25 from 81 to get 56.

$$4x^{2}+20x=56$$

Divide both sides by 4.

$$\frac{4x^{2}+20x}{4}=\frac{56}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{20}{4}x=\frac{56}{4}$$

Divide 20 by 4.

$$x^{2}+5x=\frac{56}{4}$$

Divide 56 by 4.

$$x^{2}+5x=14$$

Divide 5, the coefficient of the x term, by 2 to get \frac{5}{2}. Then add the square of \frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=14+\left(\frac{5}{2}\right)^{2}$$

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+5x+\frac{25}{4}=14+\frac{25}{4}$$

Add 14 to \frac{25}{4}.

$$x^{2}+5x+\frac{25}{4}=\frac{81}{4}$$

Factor x^{2}+5x+\frac{25}{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}.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{81}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{81}{4}}$$

Simplify.

$$x+\frac{5}{2}=\frac{9}{2} x+\frac{5}{2}=-\frac{9}{2}$$

Subtract \frac{5}{2} from both sides of the equation.

$$x=2 x=-7$$