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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-4y^2-3y+4y+3)=0

We get rid of parentheses

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

We add all the numbers together, and all the variables

-4y^2+y+3=0

a = -4; b = 1; c = +3;
Δ = b2-4ac
Δ = 12-4·(-4)·3
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-7}{2*-4}=\frac{-8}{-8}
=1
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+7}{2*-4}=\frac{6}{-8}
=-3/4
$