16(4-3m)=(-m/2+)

Solution for 16(4-3m)=(-m/2+) equation:

16(4-3m)=(-m/2+)

We move all terms to the left:

16(4-3m)-((-m/2+))=0

We add all the numbers together, and all the variables

16(-3m+4)-((-m/2))=0

We multiply parentheses

-48m-((-m/2))+64=0

We multiply all the terms by the denominator


-48m*2))-((-m+64*2))=0

We add all the numbers together, and all the variables

-48m*2))-((-1m+128))=0

We add all the numbers together, and all the variables

-1m-48m*2))-((=0

Wy multiply elements

-96m^2-1m=0

a = -96; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·(-96)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*-96}=\frac{0}{-192}
=0
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*-96}=\frac{2}{-192}
=-1/96
$