5^2+b^2=400

Solution for 5^2+b^2=400 equation:

5^2+b^2=400

We move all terms to the left:

5^2+b^2-(400)=0

determiningTheFunctionDomain
b^2-400+5^2=0

We add all the numbers together, and all the variables

b^2-375=0

a = 1; b = 0; c = -375;
Δ = b2-4ac
Δ = 02-4·1·(-375)
Δ = 1500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1500}=\sqrt{100*15}=\sqrt{100}*\sqrt{15}=10\sqrt{15}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{15}}{2*1}=\frac{0-10\sqrt{15}}{2}
=-\frac{10\sqrt{15}}{2}
=-5\sqrt{15}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{15}}{2*1}=\frac{0+10\sqrt{15}}{2}
=\frac{10\sqrt{15}}{2}
=5\sqrt{15}
$