3/4y+6=5y+3

Solution for 3/4y+6=5y+3 equation:

3/4y+6=5y+3

We move all terms to the left:

3/4y+6-(5y+3)=0


Domain of the equation: 4y!=0

y!=0/4

y!=0

y∈R


We get rid of parentheses

3/4y-5y-3+6=0

We multiply all the terms by the denominator


-5y*4y-3*4y+6*4y+3=0

Wy multiply elements

-20y^2-12y+24y+3=0

We add all the numbers together, and all the variables

-20y^2+12y+3=0

a = -20; b = 12; c = +3;
Δ = b2-4ac
Δ = 122-4·(-20)·3
Δ = 384
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8\sqrt{6}}{2*-20}=\frac{-12-8\sqrt{6}}{-40}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8\sqrt{6}}{2*-20}=\frac{-12+8\sqrt{6}}{-40}
$