-3y+y2=260

Solution for -3y+y2=260 equation:

-3y+y2=260

We move all terms to the left:

-3y+y2-(260)=0

We add all the numbers together, and all the variables

y^2-3y-260=0

a = 1; b = -3; c = -260;
Δ = b2-4ac
Δ = -32-4·1·(-260)
Δ = 1049
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{1049}}{2*1}=\frac{3-\sqrt{1049}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{1049}}{2*1}=\frac{3+\sqrt{1049}}{2}
$