2a^2+20a+23=0

Solution for 2a^2+20a+23=0 equation:

2a^2+20a+23=0

a = 2; b = 20; c = +23;
Δ = b2-4ac
Δ = 202-4·2·23
Δ = 216
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:

$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-6\sqrt{6}}{2*2}=\frac{-20-6\sqrt{6}}{4}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+6\sqrt{6}}{2*2}=\frac{-20+6\sqrt{6}}{4}
$