Solve 2x^2-4=5/3+x^2

$$2 x ^ { 2 } - 4 = \frac { 5 } { 3 } + x ^ { 2 }$$

Solve for x

$$x = \frac{\sqrt{51}}{3} \approx 2.380476143$$

$$x = -\frac{\sqrt{51}}{3} \approx -2.380476143$$

Steps by Finding Square Root

Subtract x^{2} from both sides.

$$2x^{2}-4-x^{2}=\frac{5}{3}$$

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

$$x^{2}-4=\frac{5}{3}$$

Add 4 to both sides.

$$x^{2}=\frac{5}{3}+4$$

Add \frac{5}{3} and 4 to get \frac{17}{3}.

$$x^{2}=\frac{17}{3}$$

Take the square root of both sides of the equation.

$$x=\frac{\sqrt{51}}{3} x=-\frac{\sqrt{51}}{3}$$

Steps Using the Quadratic Formula

Subtract \frac{5}{3} from both sides.

$$2x^{2}-4-\frac{5}{3}=x^{2}$$

Subtract \frac{5}{3} from -4 to get -\frac{17}{3}.

$$2x^{2}-\frac{17}{3}=x^{2}$$

Subtract x^{2} from both sides.

$$2x^{2}-\frac{17}{3}-x^{2}=0$$

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

$$x^{2}-\frac{17}{3}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -\frac{17}{3} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-\frac{17}{3}\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-\frac{17}{3}\right)}}{2}$$

Multiply -4 times -\frac{17}{3}.

$$x=\frac{0±\sqrt{\frac{68}{3}}}{2}$$

Take the square root of \frac{68}{3}.

$$x=\frac{0±\frac{2\sqrt{51}}{3}}{2}$$

Now solve the equation x=\frac{0±\frac{2\sqrt{51}}{3}}{2} when ± is plus.

$$x=\frac{\sqrt{51}}{3}$$

Now solve the equation x=\frac{0±\frac{2\sqrt{51}}{3}}{2} when ± is minus.

$$x=-\frac{\sqrt{51}}{3}$$

The equation is now solved.

$$x=\frac{\sqrt{51}}{3} x=-\frac{\sqrt{51}}{3}$$