Solve [28-(17^{2}h^2)]^2+h^2=25^2

$$[ 28 - ( 17 ^ { 2 } h ^ { 2 } ) ] ^ { 2 } + h ^ { 2 } = 25 ^ { 2 }$$

Solve for h

$$h=-\frac{\sqrt{32366-22\sqrt{1725373}}}{578}\approx -0.101888867$$

$$h=\frac{\sqrt{32366-22\sqrt{1725373}}}{578}\approx 0.101888867$$

$$h=\frac{\sqrt{22\sqrt{1725373}+32366}}{578}\approx 0.42822694$$

$$h=-\frac{\sqrt{22\sqrt{1725373}+32366}}{578}\approx -0.42822694$$

Steps Using the Quadratic Formula

Calculate 17 to the power of 2 and get 289.

$$\left(28-289h^{2}\right)^{2}+h^{2}=25^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(28-289h^{2}\right)^{2}.

$$784-16184h^{2}+83521\left(h^{2}\right)^{2}+h^{2}=25^{2}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$784-16184h^{2}+83521h^{4}+h^{2}=25^{2}$$

Combine -16184h^{2} and h^{2} to get -16183h^{2}.

$$784-16183h^{2}+83521h^{4}=25^{2}$$

Calculate 25 to the power of 2 and get 625.

$$784-16183h^{2}+83521h^{4}=625$$

Subtract 625 from both sides.

$$784-16183h^{2}+83521h^{4}-625=0$$

Subtract 625 from 784 to get 159.

$$159-16183h^{2}+83521h^{4}=0$$

Substitute t for h^{2}.

$$83521t^{2}-16183t+159=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 83521 for a, -16183 for b, and 159 for c in the quadratic formula.

$$t=\frac{-\left(-16183\right)±\sqrt{\left(-16183\right)^{2}-4\times 83521\times 159}}{2\times 83521}$$

Do the calculations.

$$t=\frac{16183±11\sqrt{1725373}}{167042}$$

Solve the equation t=\frac{16183±11\sqrt{1725373}}{167042} when ± is plus and when ± is minus.

$$t=\frac{11\sqrt{1725373}+16183}{167042} t=\frac{16183-11\sqrt{1725373}}{167042}$$

Since h=t^{2}, the solutions are obtained by evaluating h=±\sqrt{t} for each t.

$$h=\frac{\sqrt{22\sqrt{1725373}+32366}}{578} h=-\frac{\sqrt{22\sqrt{1725373}+32366}}{578} h=\frac{\sqrt{32366-22\sqrt{1725373}}}{578} h=-\frac{\sqrt{32366-22\sqrt{1725373}}}{578}$$