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

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

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

We multiply parentheses ..

(+4y^2+3y+16y+12)=0

We get rid of parentheses

4y^2+3y+16y+12=0

We add all the numbers together, and all the variables

4y^2+19y+12=0

a = 4; b = 19; c = +12;
Δ = b2-4ac
Δ = 192-4·4·12
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-13}{2*4}=\frac{-32}{8}
=-4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+13}{2*4}=\frac{-6}{8}
=-3/4
$