2(m+25)=4(m-0.5)m=

Solution for 2(m+25)=4(m-0.5)m= equation:

2(m+25)=4(m-0.5)m=

We move all terms to the left:

2(m+25)-(4(m-0.5)m)=0

We multiply parentheses

2m-(4(m-0.5)m)+50=0


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

4(m-0.5)m

We multiply parentheses

4m^2-2m

Back to the equation:

-(4m^2-2m)


We get rid of parentheses

-4m^2+2m+2m+50=0

We add all the numbers together, and all the variables

-4m^2+4m+50=0

a = -4; b = 4; c = +50;
Δ = b2-4ac
Δ = 42-4·(-4)·50
Δ = 816
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:

$\sqrt{\Delta}=\sqrt{816}=\sqrt{16*51}=\sqrt{16}*\sqrt{51}=4\sqrt{51}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{51}}{2*-4}=\frac{-4-4\sqrt{51}}{-8}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{51}}{2*-4}=\frac{-4+4\sqrt{51}}{-8}
$