8m^2+41m=-36

Solution for 8m^2+41m=-36 equation:

8m^2+41m=-36

We move all terms to the left:

8m^2+41m-(-36)=0

We add all the numbers together, and all the variables

8m^2+41m+36=0

a = 8; b = 41; c = +36;
Δ = b2-4ac
Δ = 412-4·8·36
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-23}{2*8}=\frac{-64}{16}
=-4
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+23}{2*8}=\frac{-18}{16}
=-1+1/8
$