m2+15m+50=0

Solution for m2+15m+50=0 equation:

m2+15m+50=0

We add all the numbers together, and all the variables

m^2+15m+50=0

a = 1; b = 15; c = +50;
Δ = b2-4ac
Δ = 152-4·1·50
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-5}{2*1}=\frac{-20}{2}
=-10
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+5}{2*1}=\frac{-10}{2}
=-5
$