2a^2+20a+18=0

Solution for 2a^2+20a+18=0 equation:

2a^2+20a+18=0

a = 2; b = 20; c = +18;
Δ = b2-4ac
Δ = 202-4·2·18
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-16}{2*2}=\frac{-36}{4}
=-9
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+16}{2*2}=\frac{-4}{4}
=-1
$