Solve (4x+3)^2=225

$$( 4 x + 3 ) ^ { 2 } = 225$$

Solve for x

$$x=3$$

$$x = -\frac{9}{2} = -4\frac{1}{2} = -4.5$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4x+3\right)^{2}.

$$16x^{2}+24x+9=225$$

Subtract 225 from both sides.

$$16x^{2}+24x+9-225=0$$

Subtract 225 from 9 to get -216.

$$16x^{2}+24x-216=0$$

Divide both sides by 8.

$$2x^{2}+3x-27=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-27. To find a and b, set up a system to be solved.

$$a+b=3 ab=2\left(-27\right)=-54$$

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 -54.

$$-1,54 -2,27 -3,18 -6,9$$

Calculate the sum for each pair.

$$-1+54=53 -2+27=25 -3+18=15 -6+9=3$$

The solution is the pair that gives sum 3.

$$a=-6 b=9$$

Rewrite 2x^{2}+3x-27 as \left(2x^{2}-6x\right)+\left(9x-27\right).

$$\left(2x^{2}-6x\right)+\left(9x-27\right)$$

Factor out 2x in the first and 9 in the second group.

$$2x\left(x-3\right)+9\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(2x+9\right)$$

To find equation solutions, solve x-3=0 and 2x+9=0.

$$x=3 x=-\frac{9}{2}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4x+3\right)^{2}.

$$16x^{2}+24x+9=225$$

Subtract 225 from both sides.

$$16x^{2}+24x+9-225=0$$

Subtract 225 from 9 to get -216.

$$16x^{2}+24x-216=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, 24 for b, and -216 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-24±\sqrt{24^{2}-4\times 16\left(-216\right)}}{2\times 16}$$

Square 24.

$$x=\frac{-24±\sqrt{576-4\times 16\left(-216\right)}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{-24±\sqrt{576-64\left(-216\right)}}{2\times 16}$$

Multiply -64 times -216.

$$x=\frac{-24±\sqrt{576+13824}}{2\times 16}$$

Add 576 to 13824.

$$x=\frac{-24±\sqrt{14400}}{2\times 16}$$

Take the square root of 14400.

$$x=\frac{-24±120}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{-24±120}{32}$$

Now solve the equation x=\frac{-24±120}{32} when ± is plus. Add -24 to 120.

$$x=\frac{96}{32}$$

Divide 96 by 32.

$$x=3$$

Now solve the equation x=\frac{-24±120}{32} when ± is minus. Subtract 120 from -24.

$$x=-\frac{144}{32}$$

Reduce the fraction \frac{-144}{32} to lowest terms by extracting and canceling out 16.

$$x=-\frac{9}{2}$$

The equation is now solved.

$$x=3 x=-\frac{9}{2}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4x+3\right)^{2}.

$$16x^{2}+24x+9=225$$

Subtract 9 from both sides.

$$16x^{2}+24x=225-9$$

Subtract 9 from 225 to get 216.

$$16x^{2}+24x=216$$

Divide both sides by 16.

$$\frac{16x^{2}+24x}{16}=\frac{216}{16}$$

Dividing by 16 undoes the multiplication by 16.

$$x^{2}+\frac{24}{16}x=\frac{216}{16}$$

Reduce the fraction \frac{24}{16} to lowest terms by extracting and canceling out 8.

$$x^{2}+\frac{3}{2}x=\frac{216}{16}$$

Reduce the fraction \frac{216}{16} to lowest terms by extracting and canceling out 8.

$$x^{2}+\frac{3}{2}x=\frac{27}{2}$$

Divide \frac{3}{2}, the coefficient of the x term, by 2 to get \frac{3}{4}. Then add the square of \frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{3}{2}x+\left(\frac{3}{4}\right)^{2}=\frac{27}{2}+\left(\frac{3}{4}\right)^{2}$$

Square \frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{27}{2}+\frac{9}{16}$$

Add \frac{27}{2} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{225}{16}$$

Factor x^{2}+\frac{3}{2}x+\frac{9}{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}.

$$\left(x+\frac{3}{4}\right)^{2}=\frac{225}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{\frac{225}{16}}$$

Simplify.

$$x+\frac{3}{4}=\frac{15}{4} x+\frac{3}{4}=-\frac{15}{4}$$

Subtract \frac{3}{4} from both sides of the equation.

$$x=3 x=-\frac{9}{2}$$