Solve left(x+2right)^2-2x-5/3=3

$${ \left(x+2 \right) }^{ 2 } - \frac{ 2x-5 }{ 3 } = 3$$

Solve for x

$$x = -\frac{4}{3} = -1\frac{1}{3} \approx -1.333333333$$

$$x=-2$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 3.

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

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

$$3\left(x^{2}+4x+4\right)-\left(2x-5\right)=9$$

Use the distributive property to multiply 3 by x^{2}+4x+4.

$$3x^{2}+12x+12-\left(2x-5\right)=9$$

To find the opposite of 2x-5, find the opposite of each term.

$$3x^{2}+12x+12-2x+5=9$$

Combine 12x and -2x to get 10x.

$$3x^{2}+10x+12+5=9$$

Add 12 and 5 to get 17.

$$3x^{2}+10x+17=9$$

Subtract 9 from both sides.

$$3x^{2}+10x+17-9=0$$

Subtract 9 from 17 to get 8.

$$3x^{2}+10x+8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx+8. To find a and b, set up a system to be solved.

$$a+b=10 ab=3\times 8=24$$

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 24.

$$1,24 2,12 3,8 4,6$$

Calculate the sum for each pair.

$$1+24=25 2+12=14 3+8=11 4+6=10$$

The solution is the pair that gives sum 10.

$$a=4 b=6$$

Rewrite 3x^{2}+10x+8 as \left(3x^{2}+4x\right)+\left(6x+8\right).

$$\left(3x^{2}+4x\right)+\left(6x+8\right)$$

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

$$x\left(3x+4\right)+2\left(3x+4\right)$$

Factor out common term 3x+4 by using distributive property.

$$\left(3x+4\right)\left(x+2\right)$$

To find equation solutions, solve 3x+4=0 and x+2=0.

$$x=-\frac{4}{3} x=-2$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 3.

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

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

$$3\left(x^{2}+4x+4\right)-\left(2x-5\right)=9$$

Use the distributive property to multiply 3 by x^{2}+4x+4.

$$3x^{2}+12x+12-\left(2x-5\right)=9$$

To find the opposite of 2x-5, find the opposite of each term.

$$3x^{2}+12x+12-2x+5=9$$

Combine 12x and -2x to get 10x.

$$3x^{2}+10x+12+5=9$$

Add 12 and 5 to get 17.

$$3x^{2}+10x+17=9$$

Subtract 9 from both sides.

$$3x^{2}+10x+17-9=0$$

Subtract 9 from 17 to get 8.

$$3x^{2}+10x+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 10 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\times 3\times 8}}{2\times 3}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\times 3\times 8}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-10±\sqrt{100-12\times 8}}{2\times 3}$$

Multiply -12 times 8.

$$x=\frac{-10±\sqrt{100-96}}{2\times 3}$$

Add 100 to -96.

$$x=\frac{-10±\sqrt{4}}{2\times 3}$$

Take the square root of 4.

$$x=\frac{-10±2}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-10±2}{6}$$

Now solve the equation x=\frac{-10±2}{6} when ± is plus. Add -10 to 2.

$$x=-\frac{8}{6}$$

Reduce the fraction \frac{-8}{6} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-10±2}{6} when ± is minus. Subtract 2 from -10.

$$x=-\frac{12}{6}$$

Divide -12 by 6.

$$x=-2$$

The equation is now solved.

$$x=-\frac{4}{3} x=-2$$

Steps for Completing the Square

Multiply both sides of the equation by 3.

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

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

$$3\left(x^{2}+4x+4\right)-\left(2x-5\right)=9$$

Use the distributive property to multiply 3 by x^{2}+4x+4.

$$3x^{2}+12x+12-\left(2x-5\right)=9$$

To find the opposite of 2x-5, find the opposite of each term.

$$3x^{2}+12x+12-2x+5=9$$

Combine 12x and -2x to get 10x.

$$3x^{2}+10x+12+5=9$$

Add 12 and 5 to get 17.

$$3x^{2}+10x+17=9$$

Subtract 17 from both sides.

$$3x^{2}+10x=9-17$$

Subtract 17 from 9 to get -8.

$$3x^{2}+10x=-8$$

Divide both sides by 3.

$$\frac{3x^{2}+10x}{3}=-\frac{8}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{10}{3}x=-\frac{8}{3}$$

Divide \frac{10}{3}, the coefficient of the x term, by 2 to get \frac{5}{3}. Then add the square of \frac{5}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{10}{3}x+\left(\frac{5}{3}\right)^{2}=-\frac{8}{3}+\left(\frac{5}{3}\right)^{2}$$

Square \frac{5}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{10}{3}x+\frac{25}{9}=-\frac{8}{3}+\frac{25}{9}$$

Add -\frac{8}{3} to \frac{25}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{10}{3}x+\frac{25}{9}=\frac{1}{9}$$

Factor x^{2}+\frac{10}{3}x+\frac{25}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{3}\right)^{2}=\frac{1}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{3}\right)^{2}}=\sqrt{\frac{1}{9}}$$

Simplify.

$$x+\frac{5}{3}=\frac{1}{3} x+\frac{5}{3}=-\frac{1}{3}$$

Subtract \frac{5}{3} from both sides of the equation.

$$x=-\frac{4}{3} x=-2$$