(Y+3)^2+y^2=369

Solution for (Y+3)^2+y^2=369 equation:

(+3)^2+Y^2=369

We move all terms to the left:

(+3)^2+Y^2-(369)=0

determiningTheFunctionDomain
Y^2-369+(+3)^2=0

We add all the numbers together, and all the variables

Y^2-369+3^2=0

We add all the numbers together, and all the variables

Y^2-360=0

a = 1; b = 0; c = -360;
Δ = b2-4ac
Δ = 02-4·1·(-360)
Δ = 1440
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1440}=\sqrt{144*10}=\sqrt{144}*\sqrt{10}=12\sqrt{10}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{10}}{2*1}=\frac{0-12\sqrt{10}}{2}
=-\frac{12\sqrt{10}}{2}
=-6\sqrt{10}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{10}}{2*1}=\frac{0+12\sqrt{10}}{2}
=\frac{12\sqrt{10}}{2}
=6\sqrt{10}
$