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

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

Solve for x

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

$$x=-3$$

Steps Using Factoring By Grouping

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

$$2\left(x^{2}+6x+9\right)=x+3$$

Use the distributive property to multiply 2 by x^{2}+6x+9.

$$2x^{2}+12x+18=x+3$$

Subtract x from both sides.

$$2x^{2}+12x+18-x=3$$

Combine 12x and -x to get 11x.

$$2x^{2}+11x+18=3$$

Subtract 3 from both sides.

$$2x^{2}+11x+18-3=0$$

Subtract 3 from 18 to get 15.

$$2x^{2}+11x+15=0$$

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

$$a+b=11 ab=2\times 15=30$$

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 30.

$$1,30 2,15 3,10 5,6$$

Calculate the sum for each pair.

$$1+30=31 2+15=17 3+10=13 5+6=11$$

The solution is the pair that gives sum 11.

$$a=5 b=6$$

Rewrite 2x^{2}+11x+15 as \left(2x^{2}+5x\right)+\left(6x+15\right).

$$\left(2x^{2}+5x\right)+\left(6x+15\right)$$

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

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

Factor out common term 2x+5 by using distributive property.

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

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

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

Steps Using the Quadratic Formula

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

$$2\left(x^{2}+6x+9\right)=x+3$$

Use the distributive property to multiply 2 by x^{2}+6x+9.

$$2x^{2}+12x+18=x+3$$

Subtract x from both sides.

$$2x^{2}+12x+18-x=3$$

Combine 12x and -x to get 11x.

$$2x^{2}+11x+18=3$$

Subtract 3 from both sides.

$$2x^{2}+11x+18-3=0$$

Subtract 3 from 18 to get 15.

$$2x^{2}+11x+15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 11 for b, and 15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-11±\sqrt{11^{2}-4\times 2\times 15}}{2\times 2}$$

Square 11.

$$x=\frac{-11±\sqrt{121-4\times 2\times 15}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-11±\sqrt{121-8\times 15}}{2\times 2}$$

Multiply -8 times 15.

$$x=\frac{-11±\sqrt{121-120}}{2\times 2}$$

Add 121 to -120.

$$x=\frac{-11±\sqrt{1}}{2\times 2}$$

Take the square root of 1.

$$x=\frac{-11±1}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-11±1}{4}$$

Now solve the equation x=\frac{-11±1}{4} when ± is plus. Add -11 to 1.

$$x=-\frac{10}{4}$$

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

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

Now solve the equation x=\frac{-11±1}{4} when ± is minus. Subtract 1 from -11.

$$x=-\frac{12}{4}$$

Divide -12 by 4.

$$x=-3$$

The equation is now solved.

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

Steps for Completing the Square

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

$$2\left(x^{2}+6x+9\right)=x+3$$

Use the distributive property to multiply 2 by x^{2}+6x+9.

$$2x^{2}+12x+18=x+3$$

Subtract x from both sides.

$$2x^{2}+12x+18-x=3$$

Combine 12x and -x to get 11x.

$$2x^{2}+11x+18=3$$

Subtract 18 from both sides.

$$2x^{2}+11x=3-18$$

Subtract 18 from 3 to get -15.

$$2x^{2}+11x=-15$$

Divide both sides by 2.

$$\frac{2x^{2}+11x}{2}=-\frac{15}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{11}{2}x=-\frac{15}{2}$$

Divide \frac{11}{2}, the coefficient of the x term, by 2 to get \frac{11}{4}. Then add the square of \frac{11}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{11}{2}x+\left(\frac{11}{4}\right)^{2}=-\frac{15}{2}+\left(\frac{11}{4}\right)^{2}$$

Square \frac{11}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{11}{2}x+\frac{121}{16}=-\frac{15}{2}+\frac{121}{16}$$

Add -\frac{15}{2} to \frac{121}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{11}{2}x+\frac{121}{16}=\frac{1}{16}$$

Factor x^{2}+\frac{11}{2}x+\frac{121}{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}.

$$\left(x+\frac{11}{4}\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{11}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

$$x+\frac{11}{4}=\frac{1}{4} x+\frac{11}{4}=-\frac{1}{4}$$

Subtract \frac{11}{4} from both sides of the equation.

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