4m2+.10m-5=0

Solution for 4m2+.10m-5=0 equation:

4m^2+.10m-5=0

a = 4; b = .10; c = -5;
Δ = b2-4ac
Δ = .102-4·4·(-5)
Δ = 80.01
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(.10)-\sqrt{80.01}}{2*4}=\frac{-0.1-\sqrt{80.01}}{8}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(.10)+\sqrt{80.01}}{2*4}=\frac{-0.1+\sqrt{80.01}}{8}
$