Solve (2x+3)^2+(2x+3)-2=0

$$( 2 x + 3 ) ^ { 2 } + ( 2 x + 3 ) - 2 = 0$$

Solve for x

$$x = -\frac{5}{2} = -2\frac{1}{2} = -2.5$$

$$x=-1$$

Steps Using Factoring By Grouping

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

$$4x^{2}+12x+9+2x+3-2=0$$

Combine 12x and 2x to get 14x.

$$4x^{2}+14x+9+3-2=0$$

Add 9 and 3 to get 12.

$$4x^{2}+14x+12-2=0$$

Subtract 2 from 12 to get 10.

$$4x^{2}+14x+10=0$$

Divide both sides by 2.

$$2x^{2}+7x+5=0$$

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

$$a+b=7 ab=2\times 5=10$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 10.

$$1,10 2,5$$

Calculate the sum for each pair.

$$1+10=11 2+5=7$$

The solution is the pair that gives sum 7.

$$a=2 b=5$$

Rewrite 2x^{2}+7x+5 as \left(2x^{2}+2x\right)+\left(5x+5\right).

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

Factor out 2x in the first and 5 in the second group.

$$2x\left(x+1\right)+5\left(x+1\right)$$

Factor out common term x+1 by using distributive property.

$$\left(x+1\right)\left(2x+5\right)$$

To find equation solutions, solve x+1=0 and 2x+5=0.

$$x=-1 x=-\frac{5}{2}$$

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}.

$$4x^{2}+12x+9+2x+3-2=0$$

Combine 12x and 2x to get 14x.

$$4x^{2}+14x+9+3-2=0$$

Add 9 and 3 to get 12.

$$4x^{2}+14x+12-2=0$$

Subtract 2 from 12 to get 10.

$$4x^{2}+14x+10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 14 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-14±\sqrt{14^{2}-4\times 4\times 10}}{2\times 4}$$

Square 14.

$$x=\frac{-14±\sqrt{196-4\times 4\times 10}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-14±\sqrt{196-16\times 10}}{2\times 4}$$

Multiply -16 times 10.

$$x=\frac{-14±\sqrt{196-160}}{2\times 4}$$

Add 196 to -160.

$$x=\frac{-14±\sqrt{36}}{2\times 4}$$

Take the square root of 36.

$$x=\frac{-14±6}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-14±6}{8}$$

Now solve the equation x=\frac{-14±6}{8} when ± is plus. Add -14 to 6.

$$x=-\frac{8}{8}$$

Divide -8 by 8.

$$x=-1$$

Now solve the equation x=\frac{-14±6}{8} when ± is minus. Subtract 6 from -14.

$$x=-\frac{20}{8}$$

Reduce the fraction \frac{-20}{8} to lowest terms by extracting and canceling out 4.

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

The equation is now solved.

$$x=-1 x=-\frac{5}{2}$$

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}.

$$4x^{2}+12x+9+2x+3-2=0$$

Combine 12x and 2x to get 14x.

$$4x^{2}+14x+9+3-2=0$$

Add 9 and 3 to get 12.

$$4x^{2}+14x+12-2=0$$

Subtract 2 from 12 to get 10.

$$4x^{2}+14x+10=0$$

Subtract 10 from both sides. Anything subtracted from zero gives its negation.

$$4x^{2}+14x=-10$$

Divide both sides by 4.

$$\frac{4x^{2}+14x}{4}=-\frac{10}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{14}{4}x=-\frac{10}{4}$$

Reduce the fraction \frac{14}{4} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{7}{2}x=-\frac{10}{4}$$

Reduce the fraction \frac{-10}{4} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{7}{2}x=-\frac{5}{2}$$

Divide \frac{7}{2}, the coefficient of the x term, by 2 to get \frac{7}{4}. Then add the square of \frac{7}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{7}{2}x+\left(\frac{7}{4}\right)^{2}=-\frac{5}{2}+\left(\frac{7}{4}\right)^{2}$$

Square \frac{7}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{7}{2}x+\frac{49}{16}=-\frac{5}{2}+\frac{49}{16}$$

Add -\frac{5}{2} to \frac{49}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{7}{2}x+\frac{49}{16}=\frac{9}{16}$$

Factor x^{2}+\frac{7}{2}x+\frac{49}{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}.

$$\left(x+\frac{7}{4}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

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

Subtract \frac{7}{4} from both sides of the equation.

$$x=-1 x=-\frac{5}{2}$$