121+x^2=196

Solution for 121+x^2=196 equation:

121+x^2=196

We move all terms to the left:

121+x^2-(196)=0

We add all the numbers together, and all the variables

x^2-75=0

a = 1; b = 0; c = -75;
Δ = b2-4ac
Δ = 02-4·1·(-75)
Δ = 300
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{300}=\sqrt{100*3}=\sqrt{100}*\sqrt{3}=10\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{3}}{2*1}=\frac{0-10\sqrt{3}}{2}
=-\frac{10\sqrt{3}}{2}
=-5\sqrt{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{3}}{2*1}=\frac{0+10\sqrt{3}}{2}
=\frac{10\sqrt{3}}{2}
=5\sqrt{3}
$