m^2+24m+37=0

Solution for m^2+24m+37=0 equation:

m^2+24m+37=0

a = 1; b = 24; c = +37;
Δ = b2-4ac
Δ = 242-4·1·37
Δ = 428
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:

$\sqrt{\Delta}=\sqrt{428}=\sqrt{4*107}=\sqrt{4}*\sqrt{107}=2\sqrt{107}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-2\sqrt{107}}{2*1}=\frac{-24-2\sqrt{107}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+2\sqrt{107}}{2*1}=\frac{-24+2\sqrt{107}}{2}
$