20a^2+13a+2=0

Solution for 20a^2+13a+2=0 equation:

20a^2+13a+2=0

a = 20; b = 13; c = +2;
Δ = b2-4ac
Δ = 132-4·20·2
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-3}{2*20}=\frac{-16}{40}
=-2/5
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+3}{2*20}=\frac{-10}{40}
=-1/4
$