5m^2+35m+50=0

Solution for 5m^2+35m+50=0 equation:

5m^2+35m+50=0

a = 5; b = 35; c = +50;
Δ = b2-4ac
Δ = 352-4·5·50
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-15}{2*5}=\frac{-50}{10}
=-5
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+15}{2*5}=\frac{-20}{10}
=-2
$