Solve 2=(2x+3)^2

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

Solve for x

$$x=\frac{\sqrt{2}-3}{2}\approx -0.792893219$$

$$x=\frac{-\sqrt{2}-3}{2}\approx -2.207106781$$

Steps Using the Quadratic Formula

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

$$2=4x^{2}+12x+9$$

Swap sides so that all variable terms are on the left hand side.

$$4x^{2}+12x+9=2$$

Subtract 2 from both sides.

$$4x^{2}+12x+9-2=0$$

Subtract 2 from 9 to get 7.

$$4x^{2}+12x+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 12 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\times 4\times 7}}{2\times 4}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\times 4\times 7}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-12±\sqrt{144-16\times 7}}{2\times 4}$$

Multiply -16 times 7.

$$x=\frac{-12±\sqrt{144-112}}{2\times 4}$$

Add 144 to -112.

$$x=\frac{-12±\sqrt{32}}{2\times 4}$$

Take the square root of 32.

$$x=\frac{-12±4\sqrt{2}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-12±4\sqrt{2}}{8}$$

Now solve the equation x=\frac{-12±4\sqrt{2}}{8} when ± is plus. Add -12 to 4\sqrt{2}.

$$x=\frac{4\sqrt{2}-12}{8}$$

Divide -12+4\sqrt{2} by 8.

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

Now solve the equation x=\frac{-12±4\sqrt{2}}{8} when ± is minus. Subtract 4\sqrt{2} from -12.

$$x=\frac{-4\sqrt{2}-12}{8}$$

Divide -12-4\sqrt{2} by 8.

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

The equation is now solved.

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

Steps for Completing the Square

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

$$2=4x^{2}+12x+9$$

Swap sides so that all variable terms are on the left hand side.

$$4x^{2}+12x+9=2$$

Subtract 9 from both sides.

$$4x^{2}+12x=2-9$$

Subtract 9 from 2 to get -7.

$$4x^{2}+12x=-7$$

Divide both sides by 4.

$$\frac{4x^{2}+12x}{4}=-\frac{7}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{12}{4}x=-\frac{7}{4}$$

Divide 12 by 4.

$$x^{2}+3x=-\frac{7}{4}$$

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

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=-\frac{7}{4}+\left(\frac{3}{2}\right)^{2}$$

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=\frac{-7+9}{4}$$

Add -\frac{7}{4} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+3x+\frac{9}{4}=\frac{1}{2}$$

Factor x^{2}+3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{1}{2}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{1}{2}}$$

Simplify.

$$x+\frac{3}{2}=\frac{\sqrt{2}}{2} x+\frac{3}{2}=-\frac{\sqrt{2}}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

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