60a^2+198a+120=0

Solution for 60a^2+198a+120=0 equation:

60a^2+198a+120=0

a = 60; b = 198; c = +120;
Δ = b2-4ac
Δ = 1982-4·60·120
Δ = 10404
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{10404}=102$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(198)-102}{2*60}=\frac{-300}{120}
=-2+1/2
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(198)+102}{2*60}=\frac{-96}{120}
=-4/5
$