m^2+21m=198

Solution for m^2+21m=198 equation:

m^2+21m=198

We move all terms to the left:

m^2+21m-(198)=0

a = 1; b = 21; c = -198;
Δ = b2-4ac
Δ = 212-4·1·(-198)
Δ = 1233
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:

$\sqrt{\Delta}=\sqrt{1233}=\sqrt{9*137}=\sqrt{9}*\sqrt{137}=3\sqrt{137}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-3\sqrt{137}}{2*1}=\frac{-21-3\sqrt{137}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+3\sqrt{137}}{2*1}=\frac{-21+3\sqrt{137}}{2}
$