X^2+x=435

Solution for X^2+x=435 equation:

X^2+X=435

We move all terms to the left:

X^2+X-(435)=0

a = 1; b = 1; c = -435;
Δ = b2-4ac
Δ = 12-4·1·(-435)
Δ = 1741
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{1741}}{2*1}=\frac{-1-\sqrt{1741}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{1741}}{2*1}=\frac{-1+\sqrt{1741}}{2}
$