Solve 1-(3x+2)^2=0

$$1 - ( 3 x + 2 ) ^ { 2 } = 0$$

Solve for x

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

$$x=-1$$

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}.

$$1-\left(9x^{2}+12x+4\right)=0$$

To find the opposite of 9x^{2}+12x+4, find the opposite of each term.

$$1-9x^{2}-12x-4=0$$

Subtract 4 from 1 to get -3.

$$-3-9x^{2}-12x=0$$

Divide both sides by 3.

$$-1-3x^{2}-4x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-3x^{2}-4x-1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=-4 ab=-3\left(-1\right)=3$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-1 b=-3$$

Rewrite -3x^{2}-4x-1 as \left(-3x^{2}-x\right)+\left(-3x-1\right).

$$\left(-3x^{2}-x\right)+\left(-3x-1\right)$$

Factor out -x in the first and -1 in the second group.

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

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

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

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

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

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}.

$$1-\left(9x^{2}+12x+4\right)=0$$

To find the opposite of 9x^{2}+12x+4, find the opposite of each term.

$$1-9x^{2}-12x-4=0$$

Subtract 4 from 1 to get -3.

$$-3-9x^{2}-12x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-9x^{2}-12x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, -12 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\left(-9\right)\left(-3\right)}}{2\left(-9\right)}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\left(-9\right)\left(-3\right)}}{2\left(-9\right)}$$

Multiply -4 times -9.

$$x=\frac{-\left(-12\right)±\sqrt{144+36\left(-3\right)}}{2\left(-9\right)}$$

Multiply 36 times -3.

$$x=\frac{-\left(-12\right)±\sqrt{144-108}}{2\left(-9\right)}$$

Add 144 to -108.

$$x=\frac{-\left(-12\right)±\sqrt{36}}{2\left(-9\right)}$$

Take the square root of 36.

$$x=\frac{-\left(-12\right)±6}{2\left(-9\right)}$$

The opposite of -12 is 12.

$$x=\frac{12±6}{2\left(-9\right)}$$

Multiply 2 times -9.

$$x=\frac{12±6}{-18}$$

Now solve the equation x=\frac{12±6}{-18} when ± is plus. Add 12 to 6.

$$x=\frac{18}{-18}$$

Divide 18 by -18.

$$x=-1$$

Now solve the equation x=\frac{12±6}{-18} when ± is minus. Subtract 6 from 12.

$$x=\frac{6}{-18}$$

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

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

The equation is now solved.

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

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}.

$$1-\left(9x^{2}+12x+4\right)=0$$

To find the opposite of 9x^{2}+12x+4, find the opposite of each term.

$$1-9x^{2}-12x-4=0$$

Subtract 4 from 1 to get -3.

$$-3-9x^{2}-12x=0$$

Add 3 to both sides. Anything plus zero gives itself.

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

Divide both sides by -9.

$$\frac{-9x^{2}-12x}{-9}=\frac{3}{-9}$$

Dividing by -9 undoes the multiplication by -9.

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

Reduce the fraction \frac{-12}{-9} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{4}{3}x=\frac{3}{-9}$$

Reduce the fraction \frac{3}{-9} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{4}{3}x=-\frac{1}{3}$$

Divide \frac{4}{3}, the coefficient of the x term, by 2 to get \frac{2}{3}. Then add the square of \frac{2}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{4}{3}x+\left(\frac{2}{3}\right)^{2}=-\frac{1}{3}+\left(\frac{2}{3}\right)^{2}$$

Square \frac{2}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=-\frac{1}{3}+\frac{4}{9}$$

Add -\frac{1}{3} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=\frac{1}{9}$$

Factor x^{2}+\frac{4}{3}x+\frac{4}{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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract \frac{2}{3} from both sides of the equation.

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