y^2+2y+3y^2=

Solution for y^2+2y+3y^2= equation:

y^2+2y+3y^2=

We move all terms to the left:

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

We add all the numbers together, and all the variables

4y^2+2y=0

a = 4; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·4·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*4}=\frac{-4}{8}
=-1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*4}=\frac{0}{8}
=0
$