Solve (2x+1)^2-3=13

$$( 2 x + 1 ) ^ { 2 } - 3 = 13$$

Solve for x

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

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

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+1\right)^{2}.

$$4x^{2}+4x+1-3=13$$

Subtract 3 from 1 to get -2.

$$4x^{2}+4x-2=13$$

Subtract 13 from both sides.

$$4x^{2}+4x-2-13=0$$

Subtract 13 from -2 to get -15.

$$4x^{2}+4x-15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx-15. To find a and b, set up a system to be solved.

$$a+b=4 ab=4\left(-15\right)=-60$$

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 -60.

$$-1,60 -2,30 -3,20 -4,15 -5,12 -6,10$$

Calculate the sum for each pair.

$$-1+60=59 -2+30=28 -3+20=17 -4+15=11 -5+12=7 -6+10=4$$

The solution is the pair that gives sum 4.

$$a=-6 b=10$$

Rewrite 4x^{2}+4x-15 as \left(4x^{2}-6x\right)+\left(10x-15\right).

$$\left(4x^{2}-6x\right)+\left(10x-15\right)$$

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

$$2x\left(2x-3\right)+5\left(2x-3\right)$$

Factor out common term 2x-3 by using distributive property.

$$\left(2x-3\right)\left(2x+5\right)$$

To find equation solutions, solve 2x-3=0 and 2x+5=0.

$$x=\frac{3}{2} x=-\frac{5}{2}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+1\right)^{2}.

$$4x^{2}+4x+1-3=13$$

Subtract 3 from 1 to get -2.

$$4x^{2}+4x-2=13$$

Subtract 13 from both sides.

$$4x^{2}+4x-2-13=0$$

Subtract 13 from -2 to get -15.

$$4x^{2}+4x-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 4 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\times 4\left(-15\right)}}{2\times 4}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 4\left(-15\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-4±\sqrt{16-16\left(-15\right)}}{2\times 4}$$

Multiply -16 times -15.

$$x=\frac{-4±\sqrt{16+240}}{2\times 4}$$

Add 16 to 240.

$$x=\frac{-4±\sqrt{256}}{2\times 4}$$

Take the square root of 256.

$$x=\frac{-4±16}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-4±16}{8}$$

Now solve the equation x=\frac{-4±16}{8} when ± is plus. Add -4 to 16.

$$x=\frac{12}{8}$$

Reduce the fraction \frac{12}{8} to lowest terms by extracting and canceling out 4.

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

Now solve the equation x=\frac{-4±16}{8} when ± is minus. Subtract 16 from -4.

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

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

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

The equation is now solved.

$$x=\frac{3}{2} x=-\frac{5}{2}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+1\right)^{2}.

$$4x^{2}+4x+1-3=13$$

Subtract 3 from 1 to get -2.

$$4x^{2}+4x-2=13$$

Add 2 to both sides.

$$4x^{2}+4x=13+2$$

Add 13 and 2 to get 15.

$$4x^{2}+4x=15$$

Divide both sides by 4.

$$\frac{4x^{2}+4x}{4}=\frac{15}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{4}{4}x=\frac{15}{4}$$

Divide 4 by 4.

$$x^{2}+x=\frac{15}{4}$$

Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=\frac{15}{4}+\left(\frac{1}{2}\right)^{2}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+x+\frac{1}{4}=\frac{15+1}{4}$$

Add \frac{15}{4} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Factor x^{2}+x+\frac{1}{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}.

$$\left(x+\frac{1}{2}\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{4}$$

Simplify.

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

Subtract \frac{1}{2} from both sides of the equation.

$$x=\frac{3}{2} x=-\frac{5}{2}$$