a^2-196a+324=0

Solution for a^2-196a+324=0 equation:

a^2-196a+324=0

a = 1; b = -196; c = +324;
Δ = b2-4ac
Δ = -1962-4·1·324
Δ = 37120
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{37120}=\sqrt{256*145}=\sqrt{256}*\sqrt{145}=16\sqrt{145}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-196)-16\sqrt{145}}{2*1}=\frac{196-16\sqrt{145}}{2}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-196)+16\sqrt{145}}{2*1}=\frac{196+16\sqrt{145}}{2}
$