4m2+12m-15=0

Solution for 4m2+12m-15=0 equation:

4m^2+12m-15=0

a = 4; b = 12; c = -15;
Δ = b2-4ac
Δ = 122-4·4·(-15)
Δ = 384
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8\sqrt{6}}{2*4}=\frac{-12-8\sqrt{6}}{8}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8\sqrt{6}}{2*4}=\frac{-12+8\sqrt{6}}{8}
$