3y+(2y)^2=7

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

3y+(2y)^2=7

We move all terms to the left:

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

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

a = 2; b = 3; c = -7;
Δ = b2-4ac
Δ = 32-4·2·(-7)
Δ = 65
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{65}}{2*2}=\frac{-3-\sqrt{65}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{65}}{2*2}=\frac{-3+\sqrt{65}}{4}
$