Solve (3x+5)^2=400

$$( 3 x + 5 ) ^ { 2 } = 400$$

Solve for x

$$x=5$$

$$x = -\frac{25}{3} = -8\frac{1}{3} \approx -8.333333333$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+5\right)^{2}.

$$9x^{2}+30x+25=400$$

Subtract 400 from both sides.

$$9x^{2}+30x+25-400=0$$

Subtract 400 from 25 to get -375.

$$9x^{2}+30x-375=0$$

Divide both sides by 3.

$$3x^{2}+10x-125=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-125. To find a and b, set up a system to be solved.

$$a+b=10 ab=3\left(-125\right)=-375$$

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 -375.

$$-1,375 -3,125 -5,75 -15,25$$

Calculate the sum for each pair.

$$-1+375=374 -3+125=122 -5+75=70 -15+25=10$$

The solution is the pair that gives sum 10.

$$a=-15 b=25$$

Rewrite 3x^{2}+10x-125 as \left(3x^{2}-15x\right)+\left(25x-125\right).

$$\left(3x^{2}-15x\right)+\left(25x-125\right)$$

Factor out 3x in the first and 25 in the second group.

$$3x\left(x-5\right)+25\left(x-5\right)$$

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

$$\left(x-5\right)\left(3x+25\right)$$

To find equation solutions, solve x-5=0 and 3x+25=0.

$$x=5 x=-\frac{25}{3}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+5\right)^{2}.

$$9x^{2}+30x+25=400$$

Subtract 400 from both sides.

$$9x^{2}+30x+25-400=0$$

Subtract 400 from 25 to get -375.

$$9x^{2}+30x-375=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, 30 for b, and -375 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-30±\sqrt{30^{2}-4\times 9\left(-375\right)}}{2\times 9}$$

Square 30.

$$x=\frac{-30±\sqrt{900-4\times 9\left(-375\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-30±\sqrt{900-36\left(-375\right)}}{2\times 9}$$

Multiply -36 times -375.

$$x=\frac{-30±\sqrt{900+13500}}{2\times 9}$$

Add 900 to 13500.

$$x=\frac{-30±\sqrt{14400}}{2\times 9}$$

Take the square root of 14400.

$$x=\frac{-30±120}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{-30±120}{18}$$

Now solve the equation x=\frac{-30±120}{18} when ± is plus. Add -30 to 120.

$$x=\frac{90}{18}$$

Divide 90 by 18.

$$x=5$$

Now solve the equation x=\frac{-30±120}{18} when ± is minus. Subtract 120 from -30.

$$x=-\frac{150}{18}$$

Reduce the fraction \frac{-150}{18} to lowest terms by extracting and canceling out 6.

$$x=-\frac{25}{3}$$

The equation is now solved.

$$x=5 x=-\frac{25}{3}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+5\right)^{2}.

$$9x^{2}+30x+25=400$$

Subtract 25 from both sides.

$$9x^{2}+30x=400-25$$

Subtract 25 from 400 to get 375.

$$9x^{2}+30x=375$$

Divide both sides by 9.

$$\frac{9x^{2}+30x}{9}=\frac{375}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\frac{30}{9}x=\frac{375}{9}$$

Reduce the fraction \frac{30}{9} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{10}{3}x=\frac{375}{9}$$

Reduce the fraction \frac{375}{9} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{10}{3}x=\frac{125}{3}$$

Divide \frac{10}{3}, the coefficient of the x term, by 2 to get \frac{5}{3}. Then add the square of \frac{5}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{10}{3}x+\left(\frac{5}{3}\right)^{2}=\frac{125}{3}+\left(\frac{5}{3}\right)^{2}$$

Square \frac{5}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{10}{3}x+\frac{25}{9}=\frac{125}{3}+\frac{25}{9}$$

Add \frac{125}{3} to \frac{25}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{10}{3}x+\frac{25}{9}=\frac{400}{9}$$

Factor x^{2}+\frac{10}{3}x+\frac{25}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{3}\right)^{2}=\frac{400}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{3}\right)^{2}}=\sqrt{\frac{400}{9}}$$

Simplify.

$$x+\frac{5}{3}=\frac{20}{3} x+\frac{5}{3}=-\frac{20}{3}$$

Subtract \frac{5}{3} from both sides of the equation.

$$x=5 x=-\frac{25}{3}$$