4y(y+3)=16

Solution for 4y(y+3)=16 equation:

4y(y+3)=16

We move all terms to the left:

4y(y+3)-(16)=0

We multiply parentheses

4y^2+12y-16=0

a = 4; b = 12; c = -16;
Δ = b2-4ac
Δ = 122-4·4·(-16)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-20}{2*4}=\frac{-32}{8}
=-4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+20}{2*4}=\frac{8}{8}
=1
$