Solve 2(3x+5)^2-10=22

$$2 ( 3 x + 5 ) ^ { 2 } - 10 = 22$$

Solve for x

$$x=-\frac{1}{3}\approx -0.333333333$$

$$x=-3$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+5\right)^{2}.

$$2\left(9x^{2}+30x+25\right)-10=22$$

Use the distributive property to multiply 2 by 9x^{2}+30x+25.

$$18x^{2}+60x+50-10=22$$

Subtract 10 from 50 to get 40.

$$18x^{2}+60x+40=22$$

Subtract 22 from both sides.

$$18x^{2}+60x+40-22=0$$

Subtract 22 from 40 to get 18.

$$18x^{2}+60x+18=0$$

Divide both sides by 6.

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

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

$$a+b=10 ab=3\times 3=9$$

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 9.

$$1,9 3,3$$

Calculate the sum for each pair.

$$1+9=10 3+3=6$$

The solution is the pair that gives sum 10.

$$a=1 b=9$$

Rewrite 3x^{2}+10x+3 as \left(3x^{2}+x\right)+\left(9x+3\right).

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

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

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

Factor out common term 3x+1 by using distributive property.

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

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

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

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+5\right)^{2}.

$$2\left(9x^{2}+30x+25\right)-10=22$$

Use the distributive property to multiply 2 by 9x^{2}+30x+25.

$$18x^{2}+60x+50-10=22$$

Subtract 10 from 50 to get 40.

$$18x^{2}+60x+40=22$$

Subtract 22 from both sides.

$$18x^{2}+60x+40-22=0$$

Subtract 22 from 40 to get 18.

$$18x^{2}+60x+18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 18 for a, 60 for b, and 18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-60±\sqrt{60^{2}-4\times 18\times 18}}{2\times 18}$$

Square 60.

$$x=\frac{-60±\sqrt{3600-4\times 18\times 18}}{2\times 18}$$

Multiply -4 times 18.

$$x=\frac{-60±\sqrt{3600-72\times 18}}{2\times 18}$$

Multiply -72 times 18.

$$x=\frac{-60±\sqrt{3600-1296}}{2\times 18}$$

Add 3600 to -1296.

$$x=\frac{-60±\sqrt{2304}}{2\times 18}$$

Take the square root of 2304.

$$x=\frac{-60±48}{2\times 18}$$

Multiply 2 times 18.

$$x=\frac{-60±48}{36}$$

Now solve the equation x=\frac{-60±48}{36} when ± is plus. Add -60 to 48.

$$x=-\frac{12}{36}$$

Reduce the fraction \frac{-12}{36} to lowest terms by extracting and canceling out 12.

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

Now solve the equation x=\frac{-60±48}{36} when ± is minus. Subtract 48 from -60.

$$x=-\frac{108}{36}$$

Divide -108 by 36.

$$x=-3$$

The equation is now solved.

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

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+5\right)^{2}.

$$2\left(9x^{2}+30x+25\right)-10=22$$

Use the distributive property to multiply 2 by 9x^{2}+30x+25.

$$18x^{2}+60x+50-10=22$$

Subtract 10 from 50 to get 40.

$$18x^{2}+60x+40=22$$

Subtract 40 from both sides.

$$18x^{2}+60x=22-40$$

Subtract 40 from 22 to get -18.

$$18x^{2}+60x=-18$$

Divide both sides by 18.

$$\frac{18x^{2}+60x}{18}=-\frac{18}{18}$$

Dividing by 18 undoes the multiplication by 18.

$$x^{2}+\frac{60}{18}x=-\frac{18}{18}$$

Reduce the fraction \frac{60}{18} to lowest terms by extracting and canceling out 6.

$$x^{2}+\frac{10}{3}x=-\frac{18}{18}$$

Divide -18 by 18.

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

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.

$$x^{2}+\frac{10}{3}x+\left(\frac{5}{3}\right)^{2}=-1+\left(\frac{5}{3}\right)^{2}$$

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

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

Add -1 to \frac{25}{9}.

$$x^{2}+\frac{10}{3}x+\frac{25}{9}=\frac{16}{9}$$

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}.

$$\left(x+\frac{5}{3}\right)^{2}=\frac{16}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{3}\right)^{2}}=\sqrt{\frac{16}{9}}$$

Simplify.

$$x+\frac{5}{3}=\frac{4}{3} x+\frac{5}{3}=-\frac{4}{3}$$

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

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