Solve 5(3x+2)^2=9x^2-4

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

Solve for x

$$x=-\frac{2}{3}\approx -0.666666667$$

$$x=-1$$

Steps Using Factoring By Grouping

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

$$5\left(9x^{2}+12x+4\right)=9x^{2}-4$$

Use the distributive property to multiply 5 by 9x^{2}+12x+4.

$$45x^{2}+60x+20=9x^{2}-4$$

Subtract 9x^{2} from both sides.

$$45x^{2}+60x+20-9x^{2}=-4$$

Combine 45x^{2} and -9x^{2} to get 36x^{2}.

$$36x^{2}+60x+20=-4$$

Add 4 to both sides.

$$36x^{2}+60x+20+4=0$$

Add 20 and 4 to get 24.

$$36x^{2}+60x+24=0$$

Divide both sides by 12.

$$3x^{2}+5x+2=0$$

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

$$a+b=5 ab=3\times 2=6$$

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 5.

$$a=2 b=3$$

Rewrite 3x^{2}+5x+2 as \left(3x^{2}+2x\right)+\left(3x+2\right).

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

Factor out x in 3x^{2}+2x.

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

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

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

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

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

Steps Using the Quadratic Formula

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

$$5\left(9x^{2}+12x+4\right)=9x^{2}-4$$

Use the distributive property to multiply 5 by 9x^{2}+12x+4.

$$45x^{2}+60x+20=9x^{2}-4$$

Subtract 9x^{2} from both sides.

$$45x^{2}+60x+20-9x^{2}=-4$$

Combine 45x^{2} and -9x^{2} to get 36x^{2}.

$$36x^{2}+60x+20=-4$$

Add 4 to both sides.

$$36x^{2}+60x+20+4=0$$

Add 20 and 4 to get 24.

$$36x^{2}+60x+24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 36 for a, 60 for b, and 24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-60±\sqrt{60^{2}-4\times 36\times 24}}{2\times 36}$$

Square 60.

$$x=\frac{-60±\sqrt{3600-4\times 36\times 24}}{2\times 36}$$

Multiply -4 times 36.

$$x=\frac{-60±\sqrt{3600-144\times 24}}{2\times 36}$$

Multiply -144 times 24.

$$x=\frac{-60±\sqrt{3600-3456}}{2\times 36}$$

Add 3600 to -3456.

$$x=\frac{-60±\sqrt{144}}{2\times 36}$$

Take the square root of 144.

$$x=\frac{-60±12}{2\times 36}$$

Multiply 2 times 36.

$$x=\frac{-60±12}{72}$$

Now solve the equation x=\frac{-60±12}{72} when ± is plus. Add -60 to 12.

$$x=-\frac{48}{72}$$

Reduce the fraction \frac{-48}{72} to lowest terms by extracting and canceling out 24.

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

Now solve the equation x=\frac{-60±12}{72} when ± is minus. Subtract 12 from -60.

$$x=-\frac{72}{72}$$

Divide -72 by 72.

$$x=-1$$

The equation is now solved.

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

Steps for Completing the Square

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

$$5\left(9x^{2}+12x+4\right)=9x^{2}-4$$

Use the distributive property to multiply 5 by 9x^{2}+12x+4.

$$45x^{2}+60x+20=9x^{2}-4$$

Subtract 9x^{2} from both sides.

$$45x^{2}+60x+20-9x^{2}=-4$$

Combine 45x^{2} and -9x^{2} to get 36x^{2}.

$$36x^{2}+60x+20=-4$$

Subtract 20 from both sides.

$$36x^{2}+60x=-4-20$$

Subtract 20 from -4 to get -24.

$$36x^{2}+60x=-24$$

Divide both sides by 36.

$$\frac{36x^{2}+60x}{36}=-\frac{24}{36}$$

Dividing by 36 undoes the multiplication by 36.

$$x^{2}+\frac{60}{36}x=-\frac{24}{36}$$

Reduce the fraction \frac{60}{36} to lowest terms by extracting and canceling out 12.

$$x^{2}+\frac{5}{3}x=-\frac{24}{36}$$

Reduce the fraction \frac{-24}{36} to lowest terms by extracting and canceling out 12.

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

Divide \frac{5}{3}, the coefficient of the x term, by 2 to get \frac{5}{6}. Then add the square of \frac{5}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{5}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{3}x+\frac{25}{36}=-\frac{2}{3}+\frac{25}{36}$$

Add -\frac{2}{3} to \frac{25}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{5}{3}x+\frac{25}{36}=\frac{1}{36}$$

Factor x^{2}+\frac{5}{3}x+\frac{25}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{6}\right)^{2}=\frac{1}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{6}\right)^{2}}=\sqrt{\frac{1}{36}}$$

Simplify.

$$x+\frac{5}{6}=\frac{1}{6} x+\frac{5}{6}=-\frac{1}{6}$$

Subtract \frac{5}{6} from both sides of the equation.

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