2y2=3y+y2+6

Solution for 2y2=3y+y2+6 equation:

2y^2=3y+y2+6

We move all terms to the left:

2y^2-(3y+y2+6)=0

We add all the numbers together, and all the variables

2y^2-(+3y+y^2+6)=0

We get rid of parentheses

2y^2-y^2-3y-6=0

We add all the numbers together, and all the variables

y^2-3y-6=0

a = 1; b = -3; c = -6;
Δ = b2-4ac
Δ = -32-4·1·(-6)
Δ = 33
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{33}}{2*1}=\frac{3-\sqrt{33}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{33}}{2*1}=\frac{3+\sqrt{33}}{2}
$