Solve 7x^2-4x=x^2+2

$$7 x ^ { 2 } - 4 x = x ^ { 2 } + 2$$

Solve for x

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

$$x=1$$

Steps Using Factoring By Grouping

Subtract x^{2} from both sides.

$$7x^{2}-4x-x^{2}=2$$

Combine 7x^{2} and -x^{2} to get 6x^{2}.

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

Subtract 2 from both sides.

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

Divide both sides by 2.

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

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.

$$a+b=-2 ab=3\left(-1\right)=-3$$

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. The only such pair is the system solution.

$$a=-3 b=1$$

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

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

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

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

Factor out common term x-1 by using distributive property.

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

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

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

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

$$7x^{2}-4x-x^{2}=2$$

Combine 7x^{2} and -x^{2} to get 6x^{2}.

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

Subtract 2 from both sides.

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

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

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

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 6\left(-2\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-4\right)±\sqrt{16-24\left(-2\right)}}{2\times 6}$$

Multiply -24 times -2.

$$x=\frac{-\left(-4\right)±\sqrt{16+48}}{2\times 6}$$

Add 16 to 48.

$$x=\frac{-\left(-4\right)±\sqrt{64}}{2\times 6}$$

Take the square root of 64.

$$x=\frac{-\left(-4\right)±8}{2\times 6}$$

The opposite of -4 is 4.

$$x=\frac{4±8}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{4±8}{12}$$

Now solve the equation x=\frac{4±8}{12} when ± is plus. Add 4 to 8.

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

Divide 12 by 12.

$$x=1$$

Now solve the equation x=\frac{4±8}{12} when ± is minus. Subtract 8 from 4.

$$x=-\frac{4}{12}$$

Reduce the fraction \frac{-4}{12} to lowest terms by extracting and canceling out 4.

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

The equation is now solved.

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

Steps for Completing the Square

Subtract x^{2} from both sides.

$$7x^{2}-4x-x^{2}=2$$

Combine 7x^{2} and -x^{2} to get 6x^{2}.

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

Divide both sides by 6.

$$\frac{6x^{2}-4x}{6}=\frac{2}{6}$$

Dividing by 6 undoes the multiplication by 6.

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

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

$$x^{2}-\frac{2}{3}x=\frac{2}{6}$$

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

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

Divide -\frac{2}{3}, the coefficient of the x term, by 2 to get -\frac{1}{3}. Then add the square of -\frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{3} by squaring both the numerator and the denominator of the fraction.

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

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

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

Factor x^{2}-\frac{2}{3}x+\frac{1}{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}.

$$\left(x-\frac{1}{3}\right)^{2}=\frac{4}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{3}\right)^{2}}=\sqrt{\frac{4}{9}}$$

Simplify.

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

Add \frac{1}{3} to both sides of the equation.

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