Solve (x)=-2x^2+2

$$( x ) = - 2 x ^ { 2 } + 2$$

Solve for x

$$x=\frac{\sqrt{17}-1}{4}\approx 0.780776406$$

$$x=\frac{-\sqrt{17}-1}{4}\approx -1.280776406$$

Steps Using the Quadratic Formula

Add 2x^{2} to both sides.

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

Subtract 2 from both sides.

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

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.

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

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

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

Square 1.

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

Multiply -4 times 2.

$$x=\frac{-1±\sqrt{1-8\left(-2\right)}}{2\times 2}$$

Multiply -8 times -2.

$$x=\frac{-1±\sqrt{1+16}}{2\times 2}$$

Add 1 to 16.

$$x=\frac{-1±\sqrt{17}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-1±\sqrt{17}}{4}$$

Now solve the equation x=\frac{-1±\sqrt{17}}{4} when ± is plus. Add -1 to \sqrt{17}.

$$x=\frac{\sqrt{17}-1}{4}$$

Now solve the equation x=\frac{-1±\sqrt{17}}{4} when ± is minus. Subtract \sqrt{17} from -1.

$$x=\frac{-\sqrt{17}-1}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{17}-1}{4} x=\frac{-\sqrt{17}-1}{4}$$

Steps for Completing the Square

Add 2x^{2} to both sides.

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

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.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide 2 by 2.

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

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=1+\frac{1}{16}$$

Add 1 to \frac{1}{16}.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{17}{16}$$

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

$$\left(x+\frac{1}{4}\right)^{2}=\frac{17}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{17}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{\sqrt{17}}{4} x+\frac{1}{4}=-\frac{\sqrt{17}}{4}$$

Subtract \frac{1}{4} from both sides of the equation.

$$x=\frac{\sqrt{17}-1}{4} x=\frac{-\sqrt{17}-1}{4}$$