(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

(y+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
$