1+24m+m^2=0

Solution for 1+24m+m^2=0 equation:

1+24m+m^2=0

a = 1; b = 24; c = +1;
Δ = b2-4ac
Δ = 242-4·1·1
Δ = 572
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:

$\sqrt{\Delta}=\sqrt{572}=\sqrt{4*143}=\sqrt{4}*\sqrt{143}=2\sqrt{143}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-2\sqrt{143}}{2*1}=\frac{-24-2\sqrt{143}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+2\sqrt{143}}{2*1}=\frac{-24+2\sqrt{143}}{2}
$