3/4y+14=5y+12

Solution for 3/4y+14=5y+12 equation:

3/4y+14=5y+12

We move all terms to the left:

3/4y+14-(5y+12)=0


Domain of the equation: 4y!=0

y!=0/4

y!=0

y∈R


We get rid of parentheses

3/4y-5y-12+14=0

We multiply all the terms by the denominator


-5y*4y-12*4y+14*4y+3=0

Wy multiply elements

-20y^2-48y+56y+3=0

We add all the numbers together, and all the variables

-20y^2+8y+3=0

a = -20; b = 8; c = +3;
Δ = b2-4ac
Δ = 82-4·(-20)·3
Δ = 304
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:

$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{19}}{2*-20}=\frac{-8-4\sqrt{19}}{-40}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{19}}{2*-20}=\frac{-8+4\sqrt{19}}{-40}
$