(3y^2+7)/2=5

Solution for (3y^2+7)/2=5 equation:

(3y^2+7)/2=5

We move all terms to the left:

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

We multiply all the terms by the denominator


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

We add all the numbers together, and all the variables

(3y^2+7)-10=0

We get rid of parentheses

3y^2+7-10=0

We add all the numbers together, and all the variables

3y^2-3=0

a = 3; b = 0; c = -3;
Δ = b2-4ac
Δ = 02-4·3·(-3)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6}{2*3}=\frac{-6}{6}
=-1
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6}{2*3}=\frac{6}{6}
=1
$