Solve 3x^2=-7x+6

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

Solve for x

$$x=-3$$

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

Steps Using Factoring By Grouping

Add 7x to both sides.

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

Subtract 6 from both sides.

$$3x^{2}+7x-6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=7 ab=3\left(-6\right)=-18$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -18.

$$-1,18 -2,9 -3,6$$

Calculate the sum for each pair.

$$-1+18=17 -2+9=7 -3+6=3$$

The solution is the pair that gives sum 7.

$$a=-2 b=9$$

Rewrite 3x^{2}+7x-6 as \left(3x^{2}-2x\right)+\left(9x-6\right).

$$\left(3x^{2}-2x\right)+\left(9x-6\right)$$

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

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

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

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

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

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

Steps Using the Quadratic Formula

Add 7x to both sides.

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

Subtract 6 from both sides.

$$3x^{2}+7x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 7 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\times 3\left(-6\right)}}{2\times 3}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\times 3\left(-6\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-7±\sqrt{49-12\left(-6\right)}}{2\times 3}$$

Multiply -12 times -6.

$$x=\frac{-7±\sqrt{49+72}}{2\times 3}$$

Add 49 to 72.

$$x=\frac{-7±\sqrt{121}}{2\times 3}$$

Take the square root of 121.

$$x=\frac{-7±11}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-7±11}{6}$$

Now solve the equation x=\frac{-7±11}{6} when ± is plus. Add -7 to 11.

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

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

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

Now solve the equation x=\frac{-7±11}{6} when ± is minus. Subtract 11 from -7.

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

Divide -18 by 6.

$$x=-3$$

The equation is now solved.

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

Steps for Completing the Square

Add 7x to both sides.

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

Divide both sides by 3.

$$\frac{3x^{2}+7x}{3}=\frac{6}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{7}{3}x=\frac{6}{3}$$

Divide 6 by 3.

$$x^{2}+\frac{7}{3}x=2$$

Divide \frac{7}{3}, the coefficient of the x term, by 2 to get \frac{7}{6}. Then add the square of \frac{7}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{7}{3}x+\left(\frac{7}{6}\right)^{2}=2+\left(\frac{7}{6}\right)^{2}$$

Square \frac{7}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{7}{3}x+\frac{49}{36}=2+\frac{49}{36}$$

Add 2 to \frac{49}{36}.

$$x^{2}+\frac{7}{3}x+\frac{49}{36}=\frac{121}{36}$$

Factor x^{2}+\frac{7}{3}x+\frac{49}{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}.

$$\left(x+\frac{7}{6}\right)^{2}=\frac{121}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{6}\right)^{2}}=\sqrt{\frac{121}{36}}$$

Simplify.

$$x+\frac{7}{6}=\frac{11}{6} x+\frac{7}{6}=-\frac{11}{6}$$

Subtract \frac{7}{6} from both sides of the equation.

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