x^2+x=1300

Solution for x^2+x=1300 equation:

x^2+x=1300

We move all terms to the left:

x^2+x-(1300)=0

a = 1; b = 1; c = -1300;
Δ = b2-4ac
Δ = 12-4·1·(-1300)
Δ = 5201
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{5201}}{2*1}=\frac{-1-\sqrt{5201}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{5201}}{2*1}=\frac{-1+\sqrt{5201}}{2}
$