Solve (b-10)^2+144=13

$$( b - 10 ) ^ { 2 } + 144 = 13$$

Solve for b

$$b=-\sqrt{131}i+10\approx 10-11.445523142i$$

$$b=10+\sqrt{131}i\approx 10+11.445523142i$$

Steps by Finding Square Root

Subtracting 144 from itself leaves 0.

$$\left(b-10\right)^{2}=13-144$$

Subtract 144 from 13.

$$\left(b-10\right)^{2}=-131$$

Take the square root of both sides of the equation.

$$b-10=\sqrt{131}i b-10=-\sqrt{131}i$$

Add 10 to both sides of the equation.

$$b-10-\left(-10\right)=\sqrt{131}i-\left(-10\right) b-10-\left(-10\right)=-\sqrt{131}i-\left(-10\right)$$

Subtracting -10 from itself leaves 0.

$$b=\sqrt{131}i-\left(-10\right) b=-\sqrt{131}i-\left(-10\right)$$

Subtract -10 from i\sqrt{131}.

$$b=10+\sqrt{131}i$$

Subtract -10 from -i\sqrt{131}.

$$b=-\sqrt{131}i+10$$

The equation is now solved.

$$b=10+\sqrt{131}i b=-\sqrt{131}i+10$$