2(m+2.5)=4m(m-0.5)

Solution for 2(m+2.5)=4m(m-0.5) equation:

2(m+2.5)=4m(m-0.5)

We move all terms to the left:

2(m+2.5)-(4m(m-0.5))=0

We multiply parentheses

2m-(4m(m-0.5))+5=0


We calculate terms in parentheses: -(4m(m-0.5)), so:

4m(m-0.5)

We multiply parentheses

4m^2-2m

Back to the equation:

-(4m^2-2m)


We get rid of parentheses

-4m^2+2m+2m+5=0

We add all the numbers together, and all the variables

-4m^2+4m+5=0

a = -4; b = 4; c = +5;
Δ = b2-4ac
Δ = 42-4·(-4)·5
Δ = 96
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:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{6}}{2*-4}=\frac{-4-4\sqrt{6}}{-8}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{6}}{2*-4}=\frac{-4+4\sqrt{6}}{-8}
$