m^2-4m-37=0

Solution for m^2-4m-37=0 equation:

m^2-4m-37=0

a = 1; b = -4; c = -37;
Δ = b2-4ac
Δ = -42-4·1·(-37)
Δ = 164
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:

$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{41}}{2*1}=\frac{4-2\sqrt{41}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{41}}{2*1}=\frac{4+2\sqrt{41}}{2}
$