Solve (5x+4)^2=36

$$( 5 x + 4 ) ^ { 2 } = 36$$

Solve for x

$$x=\frac{2}{5}=0.4$$

$$x=-2$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(5x+4\right)^{2}.

$$25x^{2}+40x+16=36$$

Subtract 36 from both sides.

$$25x^{2}+40x+16-36=0$$

Subtract 36 from 16 to get -20.

$$25x^{2}+40x-20=0$$

Divide both sides by 5.

$$5x^{2}+8x-4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=8 ab=5\left(-4\right)=-20$$

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 -20.

$$-1,20 -2,10 -4,5$$

Calculate the sum for each pair.

$$-1+20=19 -2+10=8 -4+5=1$$

The solution is the pair that gives sum 8.

$$a=-2 b=10$$

Rewrite 5x^{2}+8x-4 as \left(5x^{2}-2x\right)+\left(10x-4\right).

$$\left(5x^{2}-2x\right)+\left(10x-4\right)$$

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

$$x\left(5x-2\right)+2\left(5x-2\right)$$

Factor out common term 5x-2 by using distributive property.

$$\left(5x-2\right)\left(x+2\right)$$

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

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

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(5x+4\right)^{2}.

$$25x^{2}+40x+16=36$$

Subtract 36 from both sides.

$$25x^{2}+40x+16-36=0$$

Subtract 36 from 16 to get -20.

$$25x^{2}+40x-20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 25 for a, 40 for b, and -20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-40±\sqrt{40^{2}-4\times 25\left(-20\right)}}{2\times 25}$$

Square 40.

$$x=\frac{-40±\sqrt{1600-4\times 25\left(-20\right)}}{2\times 25}$$

Multiply -4 times 25.

$$x=\frac{-40±\sqrt{1600-100\left(-20\right)}}{2\times 25}$$

Multiply -100 times -20.

$$x=\frac{-40±\sqrt{1600+2000}}{2\times 25}$$

Add 1600 to 2000.

$$x=\frac{-40±\sqrt{3600}}{2\times 25}$$

Take the square root of 3600.

$$x=\frac{-40±60}{2\times 25}$$

Multiply 2 times 25.

$$x=\frac{-40±60}{50}$$

Now solve the equation x=\frac{-40±60}{50} when ± is plus. Add -40 to 60.

$$x=\frac{20}{50}$$

Reduce the fraction \frac{20}{50} to lowest terms by extracting and canceling out 10.

$$x=\frac{2}{5}$$

Now solve the equation x=\frac{-40±60}{50} when ± is minus. Subtract 60 from -40.

$$x=-\frac{100}{50}$$

Divide -100 by 50.

$$x=-2$$

The equation is now solved.

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

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(5x+4\right)^{2}.

$$25x^{2}+40x+16=36$$

Subtract 16 from both sides.

$$25x^{2}+40x=36-16$$

Subtract 16 from 36 to get 20.

$$25x^{2}+40x=20$$

Divide both sides by 25.

$$\frac{25x^{2}+40x}{25}=\frac{20}{25}$$

Dividing by 25 undoes the multiplication by 25.

$$x^{2}+\frac{40}{25}x=\frac{20}{25}$$

Reduce the fraction \frac{40}{25} to lowest terms by extracting and canceling out 5.

$$x^{2}+\frac{8}{5}x=\frac{20}{25}$$

Reduce the fraction \frac{20}{25} to lowest terms by extracting and canceling out 5.

$$x^{2}+\frac{8}{5}x=\frac{4}{5}$$

Divide \frac{8}{5}, the coefficient of the x term, by 2 to get \frac{4}{5}. Then add the square of \frac{4}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{8}{5}x+\left(\frac{4}{5}\right)^{2}=\frac{4}{5}+\left(\frac{4}{5}\right)^{2}$$

Square \frac{4}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{8}{5}x+\frac{16}{25}=\frac{4}{5}+\frac{16}{25}$$

Add \frac{4}{5} to \frac{16}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{8}{5}x+\frac{16}{25}=\frac{36}{25}$$

Factor x^{2}+\frac{8}{5}x+\frac{16}{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}.

$$\left(x+\frac{4}{5}\right)^{2}=\frac{36}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{4}{5}\right)^{2}}=\sqrt{\frac{36}{25}}$$

Simplify.

$$x+\frac{4}{5}=\frac{6}{5} x+\frac{4}{5}=-\frac{6}{5}$$

Subtract \frac{4}{5} from both sides of the equation.

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