2m^2+18m-21=0

Solution for 2m^2+18m-21=0 equation:

2m^2+18m-21=0

a = 2; b = 18; c = -21;
Δ = b2-4ac
Δ = 182-4·2·(-21)
Δ = 492
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:

$\sqrt{\Delta}=\sqrt{492}=\sqrt{4*123}=\sqrt{4}*\sqrt{123}=2\sqrt{123}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{123}}{2*2}=\frac{-18-2\sqrt{123}}{4}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{123}}{2*2}=\frac{-18+2\sqrt{123}}{4}
$