m^2-8m-40=0

Solution for m^2-8m-40=0 equation:

m^2-8m-40=0

a = 1; b = -8; c = -40;
Δ = b2-4ac
Δ = -82-4·1·(-40)
Δ = 224
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:

$\sqrt{\Delta}=\sqrt{224}=\sqrt{16*14}=\sqrt{16}*\sqrt{14}=4\sqrt{14}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{14}}{2*1}=\frac{8-4\sqrt{14}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{14}}{2*1}=\frac{8+4\sqrt{14}}{2}
$