Solve (2x+8)^2-25=0

$$( 2 x + 8 ) ^ { 2 } - 25 = 0$$

Solve for x

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

$$x = -\frac{13}{2} = -6\frac{1}{2} = -6.5$$

Steps Using Factoring By Grouping

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

$$4x^{2}+32x+64-25=0$$

Subtract 25 from 64 to get 39.

$$4x^{2}+32x+39=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx+39. To find a and b, set up a system to be solved.

$$a+b=32 ab=4\times 39=156$$

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 156.

$$1,156 2,78 3,52 4,39 6,26 12,13$$

Calculate the sum for each pair.

$$1+156=157 2+78=80 3+52=55 4+39=43 6+26=32 12+13=25$$

The solution is the pair that gives sum 32.

$$a=6 b=26$$

Rewrite 4x^{2}+32x+39 as \left(4x^{2}+6x\right)+\left(26x+39\right).

$$\left(4x^{2}+6x\right)+\left(26x+39\right)$$

Factor out 2x in the first and 13 in the second group.

$$2x\left(2x+3\right)+13\left(2x+3\right)$$

Factor out common term 2x+3 by using distributive property.

$$\left(2x+3\right)\left(2x+13\right)$$

To find equation solutions, solve 2x+3=0 and 2x+13=0.

$$x=-\frac{3}{2} x=-\frac{13}{2}$$

Steps Using the Quadratic Formula

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

$$4x^{2}+32x+64-25=0$$

Subtract 25 from 64 to get 39.

$$4x^{2}+32x+39=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 32 for b, and 39 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-32±\sqrt{32^{2}-4\times 4\times 39}}{2\times 4}$$

Square 32.

$$x=\frac{-32±\sqrt{1024-4\times 4\times 39}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-32±\sqrt{1024-16\times 39}}{2\times 4}$$

Multiply -16 times 39.

$$x=\frac{-32±\sqrt{1024-624}}{2\times 4}$$

Add 1024 to -624.

$$x=\frac{-32±\sqrt{400}}{2\times 4}$$

Take the square root of 400.

$$x=\frac{-32±20}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-32±20}{8}$$

Now solve the equation x=\frac{-32±20}{8} when ± is plus. Add -32 to 20.

$$x=-\frac{12}{8}$$

Reduce the fraction \frac{-12}{8} to lowest terms by extracting and canceling out 4.

$$x=-\frac{3}{2}$$

Now solve the equation x=\frac{-32±20}{8} when ± is minus. Subtract 20 from -32.

$$x=-\frac{52}{8}$$

Reduce the fraction \frac{-52}{8} to lowest terms by extracting and canceling out 4.

$$x=-\frac{13}{2}$$

The equation is now solved.

$$x=-\frac{3}{2} x=-\frac{13}{2}$$

Steps for Completing the Square

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

$$4x^{2}+32x+64-25=0$$

Subtract 25 from 64 to get 39.

$$4x^{2}+32x+39=0$$

Subtract 39 from both sides. Anything subtracted from zero gives its negation.

$$4x^{2}+32x=-39$$

Divide both sides by 4.

$$\frac{4x^{2}+32x}{4}=-\frac{39}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{32}{4}x=-\frac{39}{4}$$

Divide 32 by 4.

$$x^{2}+8x=-\frac{39}{4}$$

Divide 8, the coefficient of the x term, by 2 to get 4. Then add the square of 4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+8x+4^{2}=-\frac{39}{4}+4^{2}$$

Square 4.

$$x^{2}+8x+16=-\frac{39}{4}+16$$

Add -\frac{39}{4} to 16.

$$x^{2}+8x+16=\frac{25}{4}$$

Factor x^{2}+8x+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}.

$$\left(x+4\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

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

Simplify.

$$x+4=\frac{5}{2} x+4=-\frac{5}{2}$$

Subtract 4 from both sides of the equation.

$$x=-\frac{3}{2} x=-\frac{13}{2}$$