m2+7m-18=0

Solution for m2+7m-18=0 equation:

m2+7m-18=0

We add all the numbers together, and all the variables

m^2+7m-18=0

a = 1; b = 7; c = -18;
Δ = b2-4ac
Δ = 72-4·1·(-18)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-11}{2*1}=\frac{-18}{2}
=-9
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+11}{2*1}=\frac{4}{2}
=2
$