Solve 2x^2-6x^2+6x-1=0

$$2 x ^ { 2 } - 6 x ^ { 2 } + 6 x - 1 = 0$$

Solve for x

$$x = \frac{\sqrt{5} + 3}{4} \approx 1.309016994$$

$$x=\frac{3-\sqrt{5}}{4}\approx 0.190983006$$

Steps Using the Quadratic Formula

Combine 2x^{2} and -6x^{2} to get -4x^{2}.

$$-4x^{2}+6x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 6 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-4\right)\left(-1\right)}}{2\left(-4\right)}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-4\right)\left(-1\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-6±\sqrt{36+16\left(-1\right)}}{2\left(-4\right)}$$

Multiply 16 times -1.

$$x=\frac{-6±\sqrt{36-16}}{2\left(-4\right)}$$

Add 36 to -16.

$$x=\frac{-6±\sqrt{20}}{2\left(-4\right)}$$

Take the square root of 20.

$$x=\frac{-6±2\sqrt{5}}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-6±2\sqrt{5}}{-8}$$

Now solve the equation x=\frac{-6±2\sqrt{5}}{-8} when ± is plus. Add -6 to 2\sqrt{5}.

$$x=\frac{2\sqrt{5}-6}{-8}$$

Divide -6+2\sqrt{5} by -8.

$$x=\frac{3-\sqrt{5}}{4}$$

Now solve the equation x=\frac{-6±2\sqrt{5}}{-8} when ± is minus. Subtract 2\sqrt{5} from -6.

$$x=\frac{-2\sqrt{5}-6}{-8}$$

Divide -6-2\sqrt{5} by -8.

$$x=\frac{\sqrt{5}+3}{4}$$

The equation is now solved.

$$x=\frac{3-\sqrt{5}}{4} x=\frac{\sqrt{5}+3}{4}$$

Steps for Completing the Square

Combine 2x^{2} and -6x^{2} to get -4x^{2}.

$$-4x^{2}+6x-1=0$$

Add 1 to both sides. Anything plus zero gives itself.

$$-4x^{2}+6x=1$$

Divide both sides by -4.

$$\frac{-4x^{2}+6x}{-4}=\frac{1}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\frac{6}{-4}x=\frac{1}{-4}$$

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

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

Divide 1 by -4.

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

Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=-\frac{1}{4}+\frac{9}{16}$$

Add -\frac{1}{4} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{5}{16}$$

Factor x^{2}-\frac{3}{2}x+\frac{9}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{4}\right)^{2}=\frac{5}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{5}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{\sqrt{5}}{4} x-\frac{3}{4}=-\frac{\sqrt{5}}{4}$$

Add \frac{3}{4} to both sides of the equation.

$$x=\frac{\sqrt{5}+3}{4} x=\frac{3-\sqrt{5}}{4}$$