20m(m+2)=19m-4

Solution for 20m(m+2)=19m-4 equation:

20m(m+2)=19m-4

We move all terms to the left:

20m(m+2)-(19m-4)=0

We multiply parentheses

20m^2+40m-(19m-4)=0

We get rid of parentheses

20m^2+40m-19m+4=0

We add all the numbers together, and all the variables

20m^2+21m+4=0

a = 20; b = 21; c = +4;
Δ = b2-4ac
Δ = 212-4·20·4
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-11}{2*20}=\frac{-32}{40}
=-4/5
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+11}{2*20}=\frac{-10}{40}
=-1/4
$