4m2+6m-81=0

Solution for 4m2+6m-81=0 equation:

4m^2+6m-81=0

a = 4; b = 6; c = -81;
Δ = b2-4ac
Δ = 62-4·4·(-81)
Δ = 1332
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:

$\sqrt{\Delta}=\sqrt{1332}=\sqrt{36*37}=\sqrt{36}*\sqrt{37}=6\sqrt{37}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{37}}{2*4}=\frac{-6-6\sqrt{37}}{8}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{37}}{2*4}=\frac{-6+6\sqrt{37}}{8}
$