Solve 2(x+3)^2=5(x+3)

$$2 ( x + 3 ) ^ { 2 } = 5 ( x + 3 )$$

Solve for x

$$x=-\frac{1}{2}=-0.5$$

$$x=-3$$

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}.

$$2\left(x^{2}+6x+9\right)=5\left(x+3\right)$$

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

$$2x^{2}+12x+18=5\left(x+3\right)$$

Use the distributive property to multiply 5 by x+3.

$$2x^{2}+12x+18=5x+15$$

Subtract 5x from both sides.

$$2x^{2}+12x+18-5x=15$$

Combine 12x and -5x to get 7x.

$$2x^{2}+7x+18=15$$

Subtract 15 from both sides.

$$2x^{2}+7x+18-15=0$$

Subtract 15 from 18 to get 3.

$$2x^{2}+7x+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=7 ab=2\times 3=6$$

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 6.

$$1,6 2,3$$

Calculate the sum for each pair.

$$1+6=7 2+3=5$$

The solution is the pair that gives sum 7.

$$a=1 b=6$$

Rewrite 2x^{2}+7x+3 as \left(2x^{2}+x\right)+\left(6x+3\right).

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

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

$$x\left(2x+1\right)+3\left(2x+1\right)$$

Factor out common term 2x+1 by using distributive property.

$$\left(2x+1\right)\left(x+3\right)$$

To find equation solutions, solve 2x+1=0 and x+3=0.

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

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}.

$$2\left(x^{2}+6x+9\right)=5\left(x+3\right)$$

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

$$2x^{2}+12x+18=5\left(x+3\right)$$

Use the distributive property to multiply 5 by x+3.

$$2x^{2}+12x+18=5x+15$$

Subtract 5x from both sides.

$$2x^{2}+12x+18-5x=15$$

Combine 12x and -5x to get 7x.

$$2x^{2}+7x+18=15$$

Subtract 15 from both sides.

$$2x^{2}+7x+18-15=0$$

Subtract 15 from 18 to get 3.

$$2x^{2}+7x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 7 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\times 2\times 3}}{2\times 2}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\times 2\times 3}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-7±\sqrt{49-8\times 3}}{2\times 2}$$

Multiply -8 times 3.

$$x=\frac{-7±\sqrt{49-24}}{2\times 2}$$

Add 49 to -24.

$$x=\frac{-7±\sqrt{25}}{2\times 2}$$

Take the square root of 25.

$$x=\frac{-7±5}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-7±5}{4}$$

Now solve the equation x=\frac{-7±5}{4} when ± is plus. Add -7 to 5.

$$x=-\frac{2}{4}$$

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

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

Now solve the equation x=\frac{-7±5}{4} when ± is minus. Subtract 5 from -7.

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

Divide -12 by 4.

$$x=-3$$

The equation is now solved.

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

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}.

$$2\left(x^{2}+6x+9\right)=5\left(x+3\right)$$

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

$$2x^{2}+12x+18=5\left(x+3\right)$$

Use the distributive property to multiply 5 by x+3.

$$2x^{2}+12x+18=5x+15$$

Subtract 5x from both sides.

$$2x^{2}+12x+18-5x=15$$

Combine 12x and -5x to get 7x.

$$2x^{2}+7x+18=15$$

Subtract 18 from both sides.

$$2x^{2}+7x=15-18$$

Subtract 18 from 15 to get -3.

$$2x^{2}+7x=-3$$

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

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.

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

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

$$x^{2}+\frac{7}{2}x+\frac{49}{16}=-\frac{3}{2}+\frac{49}{16}$$

Add -\frac{3}{2} to \frac{49}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{7}{2}x+\frac{49}{16}=\frac{25}{16}$$

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}.

$$\left(x+\frac{7}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$x+\frac{7}{4}=\frac{5}{4} x+\frac{7}{4}=-\frac{5}{4}$$

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

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