Solve 3x^2-6=7x

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

Solve for x

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

$$x=3$$

Steps Using Factoring By Grouping

Subtract 7x from both sides.

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

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

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

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 negative, the negative number has greater absolute value than the positive. 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.

$$a=-9 b=2$$

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

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

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

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

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

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

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

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

Steps Using the Quadratic Formula

Subtract 7x from both sides.

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

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.

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

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}.

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

Square -7.

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

Multiply -4 times 3.

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

Multiply -12 times -6.

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

Add 49 to 72.

$$x=\frac{-\left(-7\right)±\sqrt{121}}{2\times 3}$$

Take the square root of 121.

$$x=\frac{-\left(-7\right)±11}{2\times 3}$$

The opposite of -7 is 7.

$$x=\frac{7±11}{2\times 3}$$

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.

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

Divide 18 by 6.

$$x=3$$

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

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

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

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

The equation is now solved.

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

Steps for Completing the Square

Subtract 7x from both sides.

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

Add 6 to both sides. Anything plus zero gives itself.

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

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.

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

Add \frac{7}{6} to both sides of the equation.

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