Solve 25=x^2/3-2

$$25= \frac{ { x }^{ 2 } }{ 3 } -2$$

Solve for x

$$x=9$$

$$x=-9$$

Steps Using Difference of Squares

Multiply both sides of the equation by 3.

$$75=x^{2}-6$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-6=75$$

Subtract 75 from both sides.

$$x^{2}-6-75=0$$

Subtract 75 from -6 to get -81.

$$x^{2}-81=0$$

Consider x^{2}-81. Rewrite x^{2}-81 as x^{2}-9^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-9\right)\left(x+9\right)=0$$

To find equation solutions, solve x-9=0 and x+9=0.

$$x=9 x=-9$$

Steps by Finding Square Root

Multiply both sides of the equation by 3.

$$75=x^{2}-6$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-6=75$$

Add 6 to both sides.

$$x^{2}=75+6$$

Add 75 and 6 to get 81.

$$x^{2}=81$$

Take the square root of both sides of the equation.

$$x=9 x=-9$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 3.

$$75=x^{2}-6$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-6=75$$

Subtract 75 from both sides.

$$x^{2}-6-75=0$$

Subtract 75 from -6 to get -81.

$$x^{2}-81=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -81 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-81\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-81\right)}}{2}$$

Multiply -4 times -81.

$$x=\frac{0±\sqrt{324}}{2}$$

Take the square root of 324.

$$x=\frac{0±18}{2}$$

Now solve the equation x=\frac{0±18}{2} when ± is plus. Divide 18 by 2.

$$x=9$$

Now solve the equation x=\frac{0±18}{2} when ± is minus. Divide -18 by 2.

$$x=-9$$

The equation is now solved.

$$x=9 x=-9$$