2y+4y+7y=3y^2

Solution for 2y+4y+7y=3y^2 equation:

2y+4y+7y=3y^2

We move all terms to the left:

2y+4y+7y-(3y^2)=0

determiningTheFunctionDomain
-3y^2+2y+4y+7y=0

We add all the numbers together, and all the variables

-3y^2+13y=0

a = -3; b = 13; c = 0;
Δ = b2-4ac
Δ = 132-4·(-3)·0
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-13}{2*-3}=\frac{-26}{-6}
=4+1/3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+13}{2*-3}=\frac{0}{-6}
=0
$