Solve x^2-3x^2+2=0

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

Solve for x

$$x=1$$

$$x=-1$$

Steps by Finding Square Root

Combine x^{2} and -3x^{2} to get -2x^{2}.

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

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by -2.

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

Divide -2 by -2 to get 1.

$$x^{2}=1$$

Take the square root of both sides of the equation.

$$x=1 x=-1$$

Steps Using the Quadratic Formula

Combine x^{2} and -3x^{2} to get -2x^{2}.

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

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

$$x=\frac{0±\sqrt{0^{2}-4\left(-2\right)\times 2}}{2\left(-2\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-2\right)\times 2}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{0±\sqrt{8\times 2}}{2\left(-2\right)}$$

Multiply 8 times 2.

$$x=\frac{0±\sqrt{16}}{2\left(-2\right)}$$

Take the square root of 16.

$$x=\frac{0±4}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{0±4}{-4}$$

Now solve the equation x=\frac{0±4}{-4} when ± is plus. Divide 4 by -4.

$$x=-1$$

Now solve the equation x=\frac{0±4}{-4} when ± is minus. Divide -4 by -4.

$$x=1$$

The equation is now solved.

$$x=-1 x=1$$