Solve 3x^2-7x-6=-3x-2

$$3 x ^ { 2 } - 7 x - 6 = - 3 x - 2$$

Solve for x

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

$$x=2$$

Steps Using Factoring By Grouping

Add 3x to both sides.

$$3x^{2}-7x-6+3x=-2$$

Combine -7x and 3x to get -4x.

$$3x^{2}-4x-6=-2$$

Add 2 to both sides.

$$3x^{2}-4x-6+2=0$$

Add -6 and 2 to get -4.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=-4 ab=3\left(-4\right)=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -12.

$$1,-12 2,-6 3,-4$$

Calculate the sum for each pair.

$$1-12=-11 2-6=-4 3-4=-1$$

The solution is the pair that gives sum -4.

$$a=-6 b=2$$

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

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

Factor out 3x in the first and 2 in the second group.

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

Factor out common term x-2 by using distributive property.

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

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

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

Steps Using the Quadratic Formula

Add 3x to both sides.

$$3x^{2}-7x-6+3x=-2$$

Combine -7x and 3x to get -4x.

$$3x^{2}-4x-6=-2$$

Add 2 to both sides.

$$3x^{2}-4x-6+2=0$$

Add -6 and 2 to get -4.

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

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

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

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 3\left(-4\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-4\right)±\sqrt{16-12\left(-4\right)}}{2\times 3}$$

Multiply -12 times -4.

$$x=\frac{-\left(-4\right)±\sqrt{16+48}}{2\times 3}$$

Add 16 to 48.

$$x=\frac{-\left(-4\right)±\sqrt{64}}{2\times 3}$$

Take the square root of 64.

$$x=\frac{-\left(-4\right)±8}{2\times 3}$$

The opposite of -4 is 4.

$$x=\frac{4±8}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{4±8}{6}$$

Now solve the equation x=\frac{4±8}{6} when ± is plus. Add 4 to 8.

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

Divide 12 by 6.

$$x=2$$

Now solve the equation x=\frac{4±8}{6} when ± is minus. Subtract 8 from 4.

$$x=-\frac{4}{6}$$

Reduce the fraction \frac{-4}{6} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

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

Steps for Completing the Square

Add 3x to both sides.

$$3x^{2}-7x-6+3x=-2$$

Combine -7x and 3x to get -4x.

$$3x^{2}-4x-6=-2$$

Add 6 to both sides.

$$3x^{2}-4x=-2+6$$

Add -2 and 6 to get 4.

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

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

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

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.

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

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

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

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

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=\frac{16}{9}$$

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add \frac{2}{3} to both sides of the equation.

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