a^2-22a+73=0

Solution for a^2-22a+73=0 equation:

a^2-22a+73=0

a = 1; b = -22; c = +73;
Δ = b2-4ac
Δ = -222-4·1·73
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-8\sqrt{3}}{2*1}=\frac{22-8\sqrt{3}}{2}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+8\sqrt{3}}{2*1}=\frac{22+8\sqrt{3}}{2}
$