2s+7=12+3/4s

Solution for 2s+7=12+3/4s equation:

2s+7=12+3/4s

We move all terms to the left:

2s+7-(12+3/4s)=0


Domain of the equation: 4s)!=0

s!=0/1

s!=0

s∈R


We add all the numbers together, and all the variables

2s-(3/4s+12)+7=0

We get rid of parentheses

2s-3/4s-12+7=0

We multiply all the terms by the denominator


2s*4s-12*4s+7*4s-3=0

Wy multiply elements

8s^2-48s+28s-3=0

We add all the numbers together, and all the variables

8s^2-20s-3=0

a = 8; b = -20; c = -3;
Δ = b2-4ac
Δ = -202-4·8·(-3)
Δ = 496
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:

$\sqrt{\Delta}=\sqrt{496}=\sqrt{16*31}=\sqrt{16}*\sqrt{31}=4\sqrt{31}$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{31}}{2*8}=\frac{20-4\sqrt{31}}{16}
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{31}}{2*8}=\frac{20+4\sqrt{31}}{16}
$