25m^2+100m+19=0

Solution for 25m^2+100m+19=0 equation:

25m^2+100m+19=0

a = 25; b = 100; c = +19;
Δ = b2-4ac
Δ = 1002-4·25·19
Δ = 8100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{8100}=90$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(100)-90}{2*25}=\frac{-190}{50}
=-3+4/5
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(100)+90}{2*25}=\frac{-10}{50}
=-1/5
$