Solve 10=x^2+1/5

$$10 = \frac { x ^ { 2 } + 1 } { 5 }$$

Solve for x

$$x=7$$

$$x=-7$$

Steps Using Difference of Squares

Multiply both sides by 5.

$$10\times 5=x^{2}+1$$

Multiply 10 and 5 to get 50.

$$50=x^{2}+1$$

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

$$x^{2}+1=50$$

Subtract 50 from both sides.

$$x^{2}+1-50=0$$

Subtract 50 from 1 to get -49.

$$x^{2}-49=0$$

Consider x^{2}-49. Rewrite x^{2}-49 as x^{2}-7^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-7\right)\left(x+7\right)=0$$

To find equation solutions, solve x-7=0 and x+7=0.

$$x=7 x=-7$$

Steps by Finding Square Root

Multiply both sides by 5.

$$10\times 5=x^{2}+1$$

Multiply 10 and 5 to get 50.

$$50=x^{2}+1$$

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

$$x^{2}+1=50$$

Subtract 1 from both sides.

$$x^{2}=50-1$$

Subtract 1 from 50 to get 49.

$$x^{2}=49$$

Take the square root of both sides of the equation.

$$x=7 x=-7$$

Steps Using the Quadratic Formula

Multiply both sides by 5.

$$10\times 5=x^{2}+1$$

Multiply 10 and 5 to get 50.

$$50=x^{2}+1$$

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

$$x^{2}+1=50$$

Subtract 50 from both sides.

$$x^{2}+1-50=0$$

Subtract 50 from 1 to get -49.

$$x^{2}-49=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -49 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-49\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-49\right)}}{2}$$

Multiply -4 times -49.

$$x=\frac{0±\sqrt{196}}{2}$$

Take the square root of 196.

$$x=\frac{0±14}{2}$$

Now solve the equation x=\frac{0±14}{2} when ± is plus. Divide 14 by 2.

$$x=7$$

Now solve the equation x=\frac{0±14}{2} when ± is minus. Divide -14 by 2.

$$x=-7$$

The equation is now solved.

$$x=7 x=-7$$