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

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

Solve for x

$$x=\frac{\sqrt{7}-1}{2}\approx 0.822875656$$

$$x=\frac{-\sqrt{7}-1}{2}\approx -1.822875656$$

Steps Using the Quadratic Formula

Subtract 2x^{2} from both sides.

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

Subtract 4x from both sides.

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

Combine 2x and -4x to get -2x.

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

Add 3 to both sides.

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

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.

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

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

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

Square -2.

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

Multiply -4 times -2.

$$x=\frac{-\left(-2\right)±\sqrt{4+8\times 3}}{2\left(-2\right)}$$

Multiply 8 times 3.

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

Add 4 to 24.

$$x=\frac{-\left(-2\right)±\sqrt{28}}{2\left(-2\right)}$$

Take the square root of 28.

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

The opposite of -2 is 2.

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

Multiply 2 times -2.

$$x=\frac{2±2\sqrt{7}}{-4}$$

Now solve the equation x=\frac{2±2\sqrt{7}}{-4} when ± is plus. Add 2 to 2\sqrt{7}.

$$x=\frac{2\sqrt{7}+2}{-4}$$

Divide 2+2\sqrt{7} by -4.

$$x=\frac{-\sqrt{7}-1}{2}$$

Now solve the equation x=\frac{2±2\sqrt{7}}{-4} when ± is minus. Subtract 2\sqrt{7} from 2.

$$x=\frac{2-2\sqrt{7}}{-4}$$

Divide 2-2\sqrt{7} by -4.

$$x=\frac{\sqrt{7}-1}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{7}-1}{2} x=\frac{\sqrt{7}-1}{2}$$

Steps for Completing the Square

Subtract 2x^{2} from both sides.

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

Subtract 4x from both sides.

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

Combine 2x and -4x to get -2x.

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

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide -2 by -2.

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

Divide -3 by -2.

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

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

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

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

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

Add \frac{3}{2} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+x+\frac{1}{4}=\frac{7}{4}$$

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

$$\left(x+\frac{1}{2}\right)^{2}=\frac{7}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{7}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{\sqrt{7}}{2} x+\frac{1}{2}=-\frac{\sqrt{7}}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$x=\frac{\sqrt{7}-1}{2} x=\frac{-\sqrt{7}-1}{2}$$