-3m^2-15m+18=0

Solution for -3m^2-15m+18=0 equation:

-3m^2-15m+18=0

a = -3; b = -15; c = +18;
Δ = b2-4ac
Δ = -152-4·(-3)·18
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-21}{2*-3}=\frac{-6}{-6}
=1
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+21}{2*-3}=\frac{36}{-6}
=-6
$