3y2+-8y=16

Solution for 3y2+-8y=16 equation:

3y^2+-8y=16

We move all terms to the left:

3y^2+-8y-(16)=0

determiningTheFunctionDomain
3y^2-8y-16+=0

We add all the numbers together, and all the variables

3y^2-8y=0

a = 3; b = -8; c = 0;
Δ = b2-4ac
Δ = -82-4·3·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8}{2*3}=\frac{0}{6}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8}{2*3}=\frac{16}{6}
=2+2/3
$