2-2s=3/4s+12

Solution for 2-2s=3/4s+12 equation:

2-2s=3/4s+12

We move all terms to the left:

2-2s-(3/4s+12)=0


Domain of the equation: 4s+12)!=0

s∈R


We get rid of parentheses

-2s-3/4s-12+2=0

We multiply all the terms by the denominator


-2s*4s-12*4s+2*4s-3=0

Wy multiply elements

-8s^2-48s+8s-3=0

We add all the numbers together, and all the variables

-8s^2-40s-3=0

a = -8; b = -40; c = -3;
Δ = b2-4ac
Δ = -402-4·(-8)·(-3)
Δ = 1504
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1504}=\sqrt{16*94}=\sqrt{16}*\sqrt{94}=4\sqrt{94}$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-4\sqrt{94}}{2*-8}=\frac{40-4\sqrt{94}}{-16}
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+4\sqrt{94}}{2*-8}=\frac{40+4\sqrt{94}}{-16}
$