1/5y-3=8y+4

Solution for 1/5y-3=8y+4 equation:

1/5y-3=8y+4

We move all terms to the left:

1/5y-3-(8y+4)=0


Domain of the equation: 5y!=0

y!=0/5

y!=0

y∈R


We get rid of parentheses

1/5y-8y-4-3=0

We multiply all the terms by the denominator


-8y*5y-4*5y-3*5y+1=0

Wy multiply elements

-40y^2-20y-15y+1=0

We add all the numbers together, and all the variables

-40y^2-35y+1=0

a = -40; b = -35; c = +1;
Δ = b2-4ac
Δ = -352-4·(-40)·1
Δ = 1385
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-\sqrt{1385}}{2*-40}=\frac{35-\sqrt{1385}}{-80}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+\sqrt{1385}}{2*-40}=\frac{35+\sqrt{1385}}{-80}
$