2a^2-42a+99=0

Solution for 2a^2-42a+99=0 equation:

2a^2-42a+99=0

a = 2; b = -42; c = +99;
Δ = b2-4ac
Δ = -422-4·2·99
Δ = 972
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:

$\sqrt{\Delta}=\sqrt{972}=\sqrt{324*3}=\sqrt{324}*\sqrt{3}=18\sqrt{3}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-18\sqrt{3}}{2*2}=\frac{42-18\sqrt{3}}{4}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+18\sqrt{3}}{2*2}=\frac{42+18\sqrt{3}}{4}
$