Solve 7x^2-3x=4x+x^2

$$7 x ^ { 2 } - 3 x = 4 x + x ^ { 2 }$$

Solve for x

$$x = \frac{7}{6} = 1\frac{1}{6} \approx 1.166666667$$

$$x=0$$

Steps Using Factoring

Subtract 4x from both sides.

$$7x^{2}-3x-4x=x^{2}$$

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

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

Subtract x^{2} from both sides.

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

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

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

Factor out x.

$$x\left(6x-7\right)=0$$

To find equation solutions, solve x=0 and 6x-7=0.

$$x=0 x=\frac{7}{6}$$

Steps Using the Quadratic Formula

Subtract 4x from both sides.

$$7x^{2}-3x-4x=x^{2}$$

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

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

Subtract x^{2} from both sides.

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

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

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

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

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

Take the square root of \left(-7\right)^{2}.

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

The opposite of -7 is 7.

$$x=\frac{7±7}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{7±7}{12}$$

Now solve the equation x=\frac{7±7}{12} when ± is plus. Add 7 to 7.

$$x=\frac{14}{12}$$

Reduce the fraction \frac{14}{12} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{7±7}{12} when ± is minus. Subtract 7 from 7.

$$x=\frac{0}{12}$$

Divide 0 by 12.

$$x=0$$

The equation is now solved.

$$x=\frac{7}{6} x=0$$

Steps for Completing the Square

Subtract 4x from both sides.

$$7x^{2}-3x-4x=x^{2}$$

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

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

Subtract x^{2} from both sides.

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

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

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

Divide both sides by 6.

$$\frac{6x^{2}-7x}{6}=\frac{0}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}-\frac{7}{6}x=\frac{0}{6}$$

Divide 0 by 6.

$$x^{2}-\frac{7}{6}x=0$$

Divide -\frac{7}{6}, the coefficient of the x term, by 2 to get -\frac{7}{12}. Then add the square of -\frac{7}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{6}x+\left(-\frac{7}{12}\right)^{2}=\left(-\frac{7}{12}\right)^{2}$$

Square -\frac{7}{12} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{6}x+\frac{49}{144}=\frac{49}{144}$$

Factor x^{2}-\frac{7}{6}x+\frac{49}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{12}\right)^{2}=\frac{49}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{12}\right)^{2}}=\sqrt{\frac{49}{144}}$$

Simplify.

$$x-\frac{7}{12}=\frac{7}{12} x-\frac{7}{12}=-\frac{7}{12}$$

Add \frac{7}{12} to both sides of the equation.

$$x=\frac{7}{6} x=0$$