X^2+y=410

Solution for X^2+y=410 equation:

X^2+X=410

We move all terms to the left:

X^2+X-(410)=0

a = 1; b = 1; c = -410;
Δ = b2-4ac
Δ = 12-4·1·(-410)
Δ = 1641
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{-(1)-\sqrt{1641}}{2*1}=\frac{-1-\sqrt{1641}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{1641}}{2*1}=\frac{-1+\sqrt{1641}}{2}
$