4m(m-5)=2(5-m)

Solution for 4m(m-5)=2(5-m) equation:

4m(m-5)=2(5-m)

We move all terms to the left:

4m(m-5)-(2(5-m))=0

We add all the numbers together, and all the variables

4m(m-5)-(2(-1m+5))=0

We multiply parentheses

4m^2-20m-(2(-1m+5))=0


We calculate terms in parentheses: -(2(-1m+5)), so:

2(-1m+5)

We multiply parentheses

-2m+10

Back to the equation:

-(-2m+10)


We get rid of parentheses

4m^2-20m+2m-10=0

We add all the numbers together, and all the variables

4m^2-18m-10=0

a = 4; b = -18; c = -10;
Δ = b2-4ac
Δ = -182-4·4·(-10)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-22}{2*4}=\frac{-4}{8}
=-1/2
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+22}{2*4}=\frac{40}{8}
=5
$