3/4y-5=-1+y.

Solution for 3/4y-5=-1+y. equation:

3/4y-5=-1+y.

We move all terms to the left:

3/4y-5-(-1+y.)=0


Domain of the equation: 4y!=0

y!=0/4

y!=0

y∈R


We add all the numbers together, and all the variables

3/4y-(y.-1)-5=0

We get rid of parentheses

3/4y-y.+1-5=0

We multiply all the terms by the denominator


-(y.)*4y+1*4y-5*4y+3=0

We add all the numbers together, and all the variables

-(+y.)*4y+1*4y-5*4y+3=0

We multiply parentheses

-4y^2+1*4y-5*4y+3=0

Wy multiply elements

-4y^2+4y-20y+3=0

We add all the numbers together, and all the variables

-4y^2-16y+3=0

a = -4; b = -16; c = +3;
Δ = b2-4ac
Δ = -162-4·(-4)·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{-(-16)-4\sqrt{19}}{2*-4}=\frac{16-4\sqrt{19}}{-8}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{19}}{2*-4}=\frac{16+4\sqrt{19}}{-8}
$