(m+4)m=121

Solution for (m+4)m=121 equation:

(m+4)m=121

We move all terms to the left:

(m+4)m-(121)=0

We multiply parentheses

m^2+4m-121=0

a = 1; b = 4; c = -121;
Δ = b2-4ac
Δ = 42-4·1·(-121)
Δ = 500
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:

$\sqrt{\Delta}=\sqrt{500}=\sqrt{100*5}=\sqrt{100}*\sqrt{5}=10\sqrt{5}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-10\sqrt{5}}{2*1}=\frac{-4-10\sqrt{5}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+10\sqrt{5}}{2*1}=\frac{-4+10\sqrt{5}}{2}
$