4y(y-2)+3=19

Solution for 4y(y-2)+3=19 equation:

4y(y-2)+3=19

We move all terms to the left:

4y(y-2)+3-(19)=0

We add all the numbers together, and all the variables

4y(y-2)-16=0

We multiply parentheses

4y^2-8y-16=0

a = 4; b = -8; c = -16;
Δ = b2-4ac
Δ = -82-4·4·(-16)
Δ = 320
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8\sqrt{5}}{2*4}=\frac{8-8\sqrt{5}}{8}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8\sqrt{5}}{2*4}=\frac{8+8\sqrt{5}}{8}
$