Solve 2x^2+4x=x^2+x+1

$$2 x ^ { 2 } + 4 x = x ^ { 2 } + x + 1$$

Solve for x

$$x=\frac{\sqrt{13}-3}{2}\approx 0.302775638$$

$$x=\frac{-\sqrt{13}-3}{2}\approx -3.302775638$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

$$2x^{2}+4x-x^{2}=x+1$$

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

$$x^{2}+4x=x+1$$

Subtract x from both sides.

$$x^{2}+4x-x=1$$

Combine 4x and -x to get 3x.

$$x^{2}+3x=1$$

Subtract 1 from both sides.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 3 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 3.

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

Multiply -4 times -1.

$$x=\frac{-3±\sqrt{9+4}}{2}$$

Add 9 to 4.

$$x=\frac{-3±\sqrt{13}}{2}$$

Now solve the equation x=\frac{-3±\sqrt{13}}{2} when ± is plus. Add -3 to \sqrt{13}.

$$x=\frac{\sqrt{13}-3}{2}$$

Now solve the equation x=\frac{-3±\sqrt{13}}{2} when ± is minus. Subtract \sqrt{13} from -3.

$$x=\frac{-\sqrt{13}-3}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{13}-3}{2} x=\frac{-\sqrt{13}-3}{2}$$

Steps for Completing the Square

Subtract x^{2} from both sides.

$$2x^{2}+4x-x^{2}=x+1$$

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

$$x^{2}+4x=x+1$$

Subtract x from both sides.

$$x^{2}+4x-x=1$$

Combine 4x and -x to get 3x.

$$x^{2}+3x=1$$

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

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

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

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

Add 1 to \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=\frac{13}{4}$$

Factor x^{2}+3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{13}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{13}{4}}$$

Simplify.

$$x+\frac{3}{2}=\frac{\sqrt{13}}{2} x+\frac{3}{2}=-\frac{\sqrt{13}}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$x=\frac{\sqrt{13}-3}{2} x=\frac{-\sqrt{13}-3}{2}$$