30m^2+100m+80=0

Solution for 30m^2+100m+80=0 equation:

30m^2+100m+80=0

a = 30; b = 100; c = +80;
Δ = b2-4ac
Δ = 1002-4·30·80
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(100)-20}{2*30}=\frac{-120}{60}
=-2
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(100)+20}{2*30}=\frac{-80}{60}
=-1+1/3
$