19a+20=-3a^2

Solution for 19a+20=-3a^2 equation:

19a+20=-3a^2

We move all terms to the left:

19a+20-(-3a^2)=0

We get rid of parentheses

3a^2+19a+20=0

a = 3; b = 19; c = +20;
Δ = b2-4ac
Δ = 192-4·3·20
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-11}{2*3}=\frac{-30}{6}
=-5
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+11}{2*3}=\frac{-8}{6}
=-1+1/3
$