(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+7y+3=0

a = 4; b = 7; c = +3;
Δ = b2-4ac
Δ = 72-4·4·3
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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