3(-10)+10+y^2=180

Solution for 3(-10)+10+y^2=180 equation:

3(-10)+10+y^2=180

We move all terms to the left:

3(-10)+10+y^2-(180)=0

determiningTheFunctionDomain
y^2+10-180+3(-10)=0

We add all the numbers together, and all the variables

y^2-200=0

a = 1; b = 0; c = -200;
Δ = b2-4ac
Δ = 02-4·1·(-200)
Δ = 800
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:

$\sqrt{\Delta}=\sqrt{800}=\sqrt{400*2}=\sqrt{400}*\sqrt{2}=20\sqrt{2}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{2}}{2*1}=\frac{0-20\sqrt{2}}{2}
=-\frac{20\sqrt{2}}{2}
=-10\sqrt{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{2}}{2*1}=\frac{0+20\sqrt{2}}{2}
=\frac{20\sqrt{2}}{2}
=10\sqrt{2}
$