Solve (3sqrt{17})^2+4^2=h^2

$$( 3 \sqrt { 17 } ) ^ { 2 } + 4 ^ { 2 } = h ^ { 2 }$$

Solve for h

$$h=13$$

$$h=-13$$

Steps Using Difference of Squares

Expand \left(3\sqrt{17}\right)^{2}.

$$3^{2}\left(\sqrt{17}\right)^{2}+4^{2}=h^{2}$$

Calculate 3 to the power of 2 and get 9.

$$9\left(\sqrt{17}\right)^{2}+4^{2}=h^{2}$$

The square of \sqrt{17} is 17.

$$9\times 17+4^{2}=h^{2}$$

Multiply 9 and 17 to get 153.

$$153+4^{2}=h^{2}$$

Calculate 4 to the power of 2 and get 16.

$$153+16=h^{2}$$

Add 153 and 16 to get 169.

$$169=h^{2}$$

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

$$h^{2}=169$$

Subtract 169 from both sides.

$$h^{2}-169=0$$

Consider h^{2}-169. Rewrite h^{2}-169 as h^{2}-13^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(h-13\right)\left(h+13\right)=0$$

To find equation solutions, solve h-13=0 and h+13=0.

$$h=13 h=-13$$

Steps by Finding Square Root

Expand \left(3\sqrt{17}\right)^{2}.

$$3^{2}\left(\sqrt{17}\right)^{2}+4^{2}=h^{2}$$

Calculate 3 to the power of 2 and get 9.

$$9\left(\sqrt{17}\right)^{2}+4^{2}=h^{2}$$

The square of \sqrt{17} is 17.

$$9\times 17+4^{2}=h^{2}$$

Multiply 9 and 17 to get 153.

$$153+4^{2}=h^{2}$$

Calculate 4 to the power of 2 and get 16.

$$153+16=h^{2}$$

Add 153 and 16 to get 169.

$$169=h^{2}$$

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

$$h^{2}=169$$

Take the square root of both sides of the equation.

$$h=13 h=-13$$

Steps Using the Quadratic Formula

Expand \left(3\sqrt{17}\right)^{2}.

$$3^{2}\left(\sqrt{17}\right)^{2}+4^{2}=h^{2}$$

Calculate 3 to the power of 2 and get 9.

$$9\left(\sqrt{17}\right)^{2}+4^{2}=h^{2}$$

The square of \sqrt{17} is 17.

$$9\times 17+4^{2}=h^{2}$$

Multiply 9 and 17 to get 153.

$$153+4^{2}=h^{2}$$

Calculate 4 to the power of 2 and get 16.

$$153+16=h^{2}$$

Add 153 and 16 to get 169.

$$169=h^{2}$$

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

$$h^{2}=169$$

Subtract 169 from both sides.

$$h^{2}-169=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -169 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$h=\frac{0±\sqrt{0^{2}-4\left(-169\right)}}{2}$$

Square 0.

$$h=\frac{0±\sqrt{-4\left(-169\right)}}{2}$$

Multiply -4 times -169.

$$h=\frac{0±\sqrt{676}}{2}$$

Take the square root of 676.

$$h=\frac{0±26}{2}$$

Now solve the equation h=\frac{0±26}{2} when ± is plus. Divide 26 by 2.

$$h=13$$

Now solve the equation h=\frac{0±26}{2} when ± is minus. Divide -26 by 2.

$$h=-13$$

The equation is now solved.

$$h=13 h=-13$$