Solve (5x+6)^2=49

$$( 5 x + 6 ) ^ { 2 } = 49$$

Solve for x

$$x=\frac{1}{5}=0.2$$

$$x = -\frac{13}{5} = -2\frac{3}{5} = -2.6$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(5x+6\right)^{2}.

$$25x^{2}+60x+36=49$$

Subtract 49 from both sides.

$$25x^{2}+60x+36-49=0$$

Subtract 49 from 36 to get -13.

$$25x^{2}+60x-13=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 25x^{2}+ax+bx-13. To find a and b, set up a system to be solved.

$$a+b=60 ab=25\left(-13\right)=-325$$

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 -325.

$$-1,325 -5,65 -13,25$$

Calculate the sum for each pair.

$$-1+325=324 -5+65=60 -13+25=12$$

The solution is the pair that gives sum 60.

$$a=-5 b=65$$

Rewrite 25x^{2}+60x-13 as \left(25x^{2}-5x\right)+\left(65x-13\right).

$$\left(25x^{2}-5x\right)+\left(65x-13\right)$$

Factor out 5x in the first and 13 in the second group.

$$5x\left(5x-1\right)+13\left(5x-1\right)$$

Factor out common term 5x-1 by using distributive property.

$$\left(5x-1\right)\left(5x+13\right)$$

To find equation solutions, solve 5x-1=0 and 5x+13=0.

$$x=\frac{1}{5} x=-\frac{13}{5}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(5x+6\right)^{2}.

$$25x^{2}+60x+36=49$$

Subtract 49 from both sides.

$$25x^{2}+60x+36-49=0$$

Subtract 49 from 36 to get -13.

$$25x^{2}+60x-13=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 25 for a, 60 for b, and -13 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-60±\sqrt{60^{2}-4\times 25\left(-13\right)}}{2\times 25}$$

Square 60.

$$x=\frac{-60±\sqrt{3600-4\times 25\left(-13\right)}}{2\times 25}$$

Multiply -4 times 25.

$$x=\frac{-60±\sqrt{3600-100\left(-13\right)}}{2\times 25}$$

Multiply -100 times -13.

$$x=\frac{-60±\sqrt{3600+1300}}{2\times 25}$$

Add 3600 to 1300.

$$x=\frac{-60±\sqrt{4900}}{2\times 25}$$

Take the square root of 4900.

$$x=\frac{-60±70}{2\times 25}$$

Multiply 2 times 25.

$$x=\frac{-60±70}{50}$$

Now solve the equation x=\frac{-60±70}{50} when ± is plus. Add -60 to 70.

$$x=\frac{10}{50}$$

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

$$x=\frac{1}{5}$$

Now solve the equation x=\frac{-60±70}{50} when ± is minus. Subtract 70 from -60.

$$x=-\frac{130}{50}$$

Reduce the fraction \frac{-130}{50} to lowest terms by extracting and canceling out 10.

$$x=-\frac{13}{5}$$

The equation is now solved.

$$x=\frac{1}{5} x=-\frac{13}{5}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(5x+6\right)^{2}.

$$25x^{2}+60x+36=49$$

Subtract 36 from both sides.

$$25x^{2}+60x=49-36$$

Subtract 36 from 49 to get 13.

$$25x^{2}+60x=13$$

Divide both sides by 25.

$$\frac{25x^{2}+60x}{25}=\frac{13}{25}$$

Dividing by 25 undoes the multiplication by 25.

$$x^{2}+\frac{60}{25}x=\frac{13}{25}$$

Reduce the fraction \frac{60}{25} to lowest terms by extracting and canceling out 5.

$$x^{2}+\frac{12}{5}x=\frac{13}{25}$$

Divide \frac{12}{5}, the coefficient of the x term, by 2 to get \frac{6}{5}. Then add the square of \frac{6}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{12}{5}x+\left(\frac{6}{5}\right)^{2}=\frac{13}{25}+\left(\frac{6}{5}\right)^{2}$$

Square \frac{6}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{12}{5}x+\frac{36}{25}=\frac{13+36}{25}$$

Add \frac{13}{25} to \frac{36}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{12}{5}x+\frac{36}{25}=\frac{49}{25}$$

Factor x^{2}+\frac{12}{5}x+\frac{36}{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}.

$$\left(x+\frac{6}{5}\right)^{2}=\frac{49}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{6}{5}\right)^{2}}=\sqrt{\frac{49}{25}}$$

Simplify.

$$x+\frac{6}{5}=\frac{7}{5} x+\frac{6}{5}=-\frac{7}{5}$$

Subtract \frac{6}{5} from both sides of the equation.

$$x=\frac{1}{5} x=-\frac{13}{5}$$