Solve 2^5+b^2=100

$$2 ^ { 5 } + b ^ { 2 } = 100$$

Solve for b

$$b=2\sqrt{17}\approx 8.246211251$$

$$b=-2\sqrt{17}\approx -8.246211251$$

Steps by Finding Square Root

Calculate 2 to the power of 5 and get 32.

$$32+b^{2}=100$$

Subtract 32 from both sides.

$$b^{2}=100-32$$

Subtract 32 from 100 to get 68.

$$b^{2}=68$$

Take the square root of both sides of the equation.

$$b=2\sqrt{17} b=-2\sqrt{17}$$

Steps Using the Quadratic Formula

Calculate 2 to the power of 5 and get 32.

$$32+b^{2}=100$$

Subtract 100 from both sides.

$$32+b^{2}-100=0$$

Subtract 100 from 32 to get -68.

$$-68+b^{2}=0$$

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$b^{2}-68=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -68 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{0±\sqrt{0^{2}-4\left(-68\right)}}{2}$$

Square 0.

$$b=\frac{0±\sqrt{-4\left(-68\right)}}{2}$$

Multiply -4 times -68.

$$b=\frac{0±\sqrt{272}}{2}$$

Take the square root of 272.

$$b=\frac{0±4\sqrt{17}}{2}$$

Now solve the equation b=\frac{0±4\sqrt{17}}{2} when ± is plus.

$$b=2\sqrt{17}$$

Now solve the equation b=\frac{0±4\sqrt{17}}{2} when ± is minus.

$$b=-2\sqrt{17}$$

The equation is now solved.

$$b=2\sqrt{17} b=-2\sqrt{17}$$