Solve x^2-3+2=1

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

Solve for x

$$x=\sqrt{2}\approx 1.414213562$$

$$x=-\sqrt{2}\approx -1.414213562$$

Steps by Finding Square Root

Add -3 and 2 to get -1.

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

Add 1 to both sides.

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

Add 1 and 1 to get 2.

$$x^{2}=2$$

Take the square root of both sides of the equation.

$$x=\sqrt{2} x=-\sqrt{2}$$

Steps Using the Quadratic Formula

Add -3 and 2 to get -1.

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

Subtract 1 from both sides.

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

Subtract 1 from -1 to get -2.

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

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

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

Square 0.

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

Multiply -4 times -2.

$$x=\frac{0±\sqrt{8}}{2}$$

Take the square root of 8.

$$x=\frac{0±2\sqrt{2}}{2}$$

Now solve the equation x=\frac{0±2\sqrt{2}}{2} when ± is plus.

$$x=\sqrt{2}$$

Now solve the equation x=\frac{0±2\sqrt{2}}{2} when ± is minus.

$$x=-\sqrt{2}$$

The equation is now solved.

$$x=\sqrt{2} x=-\sqrt{2}$$