m^2+14m=24=0

Solution for m^2+14m=24=0 equation:

m^2+14m=24=0

We move all terms to the left:

m^2+14m-(24)=0

a = 1; b = 14; c = -24;
Δ = b2-4ac
Δ = 142-4·1·(-24)
Δ = 292
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:

$\sqrt{\Delta}=\sqrt{292}=\sqrt{4*73}=\sqrt{4}*\sqrt{73}=2\sqrt{73}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{73}}{2*1}=\frac{-14-2\sqrt{73}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{73}}{2*1}=\frac{-14+2\sqrt{73}}{2}
$