4^2+b^2=64

Solution for 4^2+b^2=64 equation:

4^2+b^2=64

We move all terms to the left:

4^2+b^2-(64)=0

determiningTheFunctionDomain
b^2-64+4^2=0

We add all the numbers together, and all the variables

b^2-48=0

a = 1; b = 0; c = -48;
Δ = b2-4ac
Δ = 02-4·1·(-48)
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{3}}{2*1}=\frac{0-8\sqrt{3}}{2}
=-\frac{8\sqrt{3}}{2}
=-4\sqrt{3}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{3}}{2*1}=\frac{0+8\sqrt{3}}{2}
=\frac{8\sqrt{3}}{2}
=4\sqrt{3}
$