3y^2-7=16y+5

Solution for 3y^2-7=16y+5 equation:

3y^2-7=16y+5

We move all terms to the left:

3y^2-7-(16y+5)=0

We get rid of parentheses

3y^2-16y-5-7=0

We add all the numbers together, and all the variables

3y^2-16y-12=0

a = 3; b = -16; c = -12;
Δ = b2-4ac
Δ = -162-4·3·(-12)
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{-(-16)-20}{2*3}=\frac{-4}{6}
=-2/3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+20}{2*3}=\frac{36}{6}
=6
$