Solve (5x+7)^2=16

$$( 5 x + 7 ) ^ { 2 } = 16$$

Solve for x

$$x=-\frac{3}{5}=-0.6$$

$$x = -\frac{11}{5} = -2\frac{1}{5} = -2.2$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(5x+7\right)^{2}.

$$25x^{2}+70x+49=16$$

Subtract 16 from both sides.

$$25x^{2}+70x+49-16=0$$

Subtract 16 from 49 to get 33.

$$25x^{2}+70x+33=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 25x^{2}+ax+bx+33. To find a and b, set up a system to be solved.

$$a+b=70 ab=25\times 33=825$$

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 825.

$$1,825 3,275 5,165 11,75 15,55 25,33$$

Calculate the sum for each pair.

$$1+825=826 3+275=278 5+165=170 11+75=86 15+55=70 25+33=58$$

The solution is the pair that gives sum 70.

$$a=15 b=55$$

Rewrite 25x^{2}+70x+33 as \left(25x^{2}+15x\right)+\left(55x+33\right).

$$\left(25x^{2}+15x\right)+\left(55x+33\right)$$

Factor out 5x in the first and 11 in the second group.

$$5x\left(5x+3\right)+11\left(5x+3\right)$$

Factor out common term 5x+3 by using distributive property.

$$\left(5x+3\right)\left(5x+11\right)$$

To find equation solutions, solve 5x+3=0 and 5x+11=0.

$$x=-\frac{3}{5} x=-\frac{11}{5}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(5x+7\right)^{2}.

$$25x^{2}+70x+49=16$$

Subtract 16 from both sides.

$$25x^{2}+70x+49-16=0$$

Subtract 16 from 49 to get 33.

$$25x^{2}+70x+33=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 25 for a, 70 for b, and 33 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-70±\sqrt{70^{2}-4\times 25\times 33}}{2\times 25}$$

Square 70.

$$x=\frac{-70±\sqrt{4900-4\times 25\times 33}}{2\times 25}$$

Multiply -4 times 25.

$$x=\frac{-70±\sqrt{4900-100\times 33}}{2\times 25}$$

Multiply -100 times 33.

$$x=\frac{-70±\sqrt{4900-3300}}{2\times 25}$$

Add 4900 to -3300.

$$x=\frac{-70±\sqrt{1600}}{2\times 25}$$

Take the square root of 1600.

$$x=\frac{-70±40}{2\times 25}$$

Multiply 2 times 25.

$$x=\frac{-70±40}{50}$$

Now solve the equation x=\frac{-70±40}{50} when ± is plus. Add -70 to 40.

$$x=-\frac{30}{50}$$

Reduce the fraction \frac{-30}{50} to lowest terms by extracting and canceling out 10.

$$x=-\frac{3}{5}$$

Now solve the equation x=\frac{-70±40}{50} when ± is minus. Subtract 40 from -70.

$$x=-\frac{110}{50}$$

Reduce the fraction \frac{-110}{50} to lowest terms by extracting and canceling out 10.

$$x=-\frac{11}{5}$$

The equation is now solved.

$$x=-\frac{3}{5} x=-\frac{11}{5}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(5x+7\right)^{2}.

$$25x^{2}+70x+49=16$$

Subtract 49 from both sides.

$$25x^{2}+70x=16-49$$

Subtract 49 from 16 to get -33.

$$25x^{2}+70x=-33$$

Divide both sides by 25.

$$\frac{25x^{2}+70x}{25}=-\frac{33}{25}$$

Dividing by 25 undoes the multiplication by 25.

$$x^{2}+\frac{70}{25}x=-\frac{33}{25}$$

Reduce the fraction \frac{70}{25} to lowest terms by extracting and canceling out 5.

$$x^{2}+\frac{14}{5}x=-\frac{33}{25}$$

Divide \frac{14}{5}, the coefficient of the x term, by 2 to get \frac{7}{5}. Then add the square of \frac{7}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{14}{5}x+\left(\frac{7}{5}\right)^{2}=-\frac{33}{25}+\left(\frac{7}{5}\right)^{2}$$

Square \frac{7}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{14}{5}x+\frac{49}{25}=\frac{-33+49}{25}$$

Add -\frac{33}{25} to \frac{49}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{14}{5}x+\frac{49}{25}=\frac{16}{25}$$

Factor x^{2}+\frac{14}{5}x+\frac{49}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{7}{5}\right)^{2}=\frac{16}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{5}\right)^{2}}=\sqrt{\frac{16}{25}}$$

Simplify.

$$x+\frac{7}{5}=\frac{4}{5} x+\frac{7}{5}=-\frac{4}{5}$$

Subtract \frac{7}{5} from both sides of the equation.

$$x=-\frac{3}{5} x=-\frac{11}{5}$$