Solve 3^2+sqrt7^2=c^2

$$3 ^ { 2 } + \sqrt { 7 } ^ { 2 } = c ^ { 2 }$$

Solve for c

$$c=4$$

$$c=-4$$

Steps Using Difference of Squares

Calculate 3 to the power of 2 and get 9.

$$9+\left(\sqrt{7}\right)^{2}=c^{2}$$

The square of \sqrt{7} is 7.

$$9+7=c^{2}$$

Add 9 and 7 to get 16.

$$16=c^{2}$$

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

$$c^{2}=16$$

Subtract 16 from both sides.

$$c^{2}-16=0$$

Consider c^{2}-16. Rewrite c^{2}-16 as c^{2}-4^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(c-4\right)\left(c+4\right)=0$$

To find equation solutions, solve c-4=0 and c+4=0.

$$c=4 c=-4$$

Steps by Finding Square Root

Calculate 3 to the power of 2 and get 9.

$$9+\left(\sqrt{7}\right)^{2}=c^{2}$$

The square of \sqrt{7} is 7.

$$9+7=c^{2}$$

Add 9 and 7 to get 16.

$$16=c^{2}$$

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

$$c^{2}=16$$

Take the square root of both sides of the equation.

$$c=4 c=-4$$

Steps Using the Quadratic Formula

Calculate 3 to the power of 2 and get 9.

$$9+\left(\sqrt{7}\right)^{2}=c^{2}$$

The square of \sqrt{7} is 7.

$$9+7=c^{2}$$

Add 9 and 7 to get 16.

$$16=c^{2}$$

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

$$c^{2}=16$$

Subtract 16 from both sides.

$$c^{2}-16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$c=\frac{0±\sqrt{0^{2}-4\left(-16\right)}}{2}$$

Square 0.

$$c=\frac{0±\sqrt{-4\left(-16\right)}}{2}$$

Multiply -4 times -16.

$$c=\frac{0±\sqrt{64}}{2}$$

Take the square root of 64.

$$c=\frac{0±8}{2}$$

Now solve the equation c=\frac{0±8}{2} when ± is plus. Divide 8 by 2.

$$c=4$$

Now solve the equation c=\frac{0±8}{2} when ± is minus. Divide -8 by 2.

$$c=-4$$

The equation is now solved.

$$c=4 c=-4$$