49+b^2=196

Solution for 49+b^2=196 equation:

49+b^2=196

We move all terms to the left:

49+b^2-(196)=0

We add all the numbers together, and all the variables

b^2-147=0

a = 1; b = 0; c = -147;
Δ = b2-4ac
Δ = 02-4·1·(-147)
Δ = 588
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:

$\sqrt{\Delta}=\sqrt{588}=\sqrt{196*3}=\sqrt{196}*\sqrt{3}=14\sqrt{3}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-14\sqrt{3}}{2*1}=\frac{0-14\sqrt{3}}{2}
=-\frac{14\sqrt{3}}{2}
=-7\sqrt{3}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+14\sqrt{3}}{2*1}=\frac{0+14\sqrt{3}}{2}
=\frac{14\sqrt{3}}{2}
=7\sqrt{3}
$