1/4y+3=-5y

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

1/4y+3=-5y

We move all terms to the left:

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


Domain of the equation: 4y!=0

y!=0/4

y!=0

y∈R


We get rid of parentheses

1/4y+5y+3=0

We multiply all the terms by the denominator


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

Wy multiply elements

20y^2+12y+1=0

a = 20; b = 12; c = +1;
Δ = b2-4ac
Δ = 122-4·20·1
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8}{2*20}=\frac{-20}{40}
=-1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8}{2*20}=\frac{-4}{40}
=-1/10
$