3m^2+57m+111=0

Solution for 3m^2+57m+111=0 equation:

3m^2+57m+111=0

a = 3; b = 57; c = +111;
Δ = b2-4ac
Δ = 572-4·3·111
Δ = 1917
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1917}=\sqrt{9*213}=\sqrt{9}*\sqrt{213}=3\sqrt{213}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(57)-3\sqrt{213}}{2*3}=\frac{-57-3\sqrt{213}}{6}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(57)+3\sqrt{213}}{2*3}=\frac{-57+3\sqrt{213}}{6}
$