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

Solution for (3y+y)(4y-1)=0 equation:

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

We add all the numbers together, and all the variables

(+4y)(4y-1)=0

We multiply parentheses ..

(+16y^2-4y)=0

We get rid of parentheses

16y^2-4y=0

a = 16; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·16·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*16}=\frac{0}{32}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*16}=\frac{8}{32}
=1/4
$