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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+4y^2+16y-1y-4)=0

We get rid of parentheses

4y^2+16y-1y-4=0

We add all the numbers together, and all the variables

4y^2+15y-4=0

a = 4; b = 15; c = -4;
Δ = b2-4ac
Δ = 152-4·4·(-4)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-17}{2*4}=\frac{-32}{8}
=-4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+17}{2*4}=\frac{2}{8}
=1/4
$