4m^2+m-121=11

Solution for 4m^2+m-121=11 equation:

4m^2+m-121=11

We move all terms to the left:

4m^2+m-121-(11)=0

We add all the numbers together, and all the variables

4m^2+m-132=0

a = 4; b = 1; c = -132;
Δ = b2-4ac
Δ = 12-4·4·(-132)
Δ = 2113
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{-(1)-\sqrt{2113}}{2*4}=\frac{-1-\sqrt{2113}}{8}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{2113}}{2*4}=\frac{-1+\sqrt{2113}}{8}
$