16+b^2=49

Solution for 16+b^2=49 equation:

16+b^2=49

We move all terms to the left:

16+b^2-(49)=0

We add all the numbers together, and all the variables

b^2-33=0

a = 1; b = 0; c = -33;
Δ = b2-4ac
Δ = 02-4·1·(-33)
Δ = 132
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:

$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{33}}{2*1}=\frac{0-2\sqrt{33}}{2}
=-\frac{2\sqrt{33}}{2}
=-\sqrt{33}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{33}}{2*1}=\frac{0+2\sqrt{33}}{2}
=\frac{2\sqrt{33}}{2}
=\sqrt{33}
$