X^2+y^2+3y=4

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

X^2+X^2+3X=4

We move all terms to the left:

X^2+X^2+3X-(4)=0

We add all the numbers together, and all the variables

2X^2+3X-4=0

a = 2; b = 3; c = -4;
Δ = b2-4ac
Δ = 32-4·2·(-4)
Δ = 41
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{41}}{2*2}=\frac{-3-\sqrt{41}}{4}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{41}}{2*2}=\frac{-3+\sqrt{41}}{4}
$