m^2+18m-19=0

Solution for m^2+18m-19=0 equation:

m^2+18m-19=0

a = 1; b = 18; c = -19;
Δ = b2-4ac
Δ = 182-4·1·(-19)
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{-(18)-20}{2*1}=\frac{-38}{2}
=-19
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+20}{2*1}=\frac{2}{2}
=1
$