4m(m+-5)=28

Solution for 4m(m+-5)=28 equation:

4m(m+-5)=28

We move all terms to the left:

4m(m+-5)-(28)=0

We add all the numbers together, and all the variables

4m(m-5)-28=0

We multiply parentheses

4m^2-20m-28=0

a = 4; b = -20; c = -28;
Δ = b2-4ac
Δ = -202-4·4·(-28)
Δ = 848
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:

$\sqrt{\Delta}=\sqrt{848}=\sqrt{16*53}=\sqrt{16}*\sqrt{53}=4\sqrt{53}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{53}}{2*4}=\frac{20-4\sqrt{53}}{8}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{53}}{2*4}=\frac{20+4\sqrt{53}}{8}
$