(x)^2+x=600

Solution for (x)^2+x=600 equation:

(x)^2+x=600

We move all terms to the left:

(x)^2+x-(600)=0

a = 1; b = 1; c = -600;
Δ = b2-4ac
Δ = 12-4·1·(-600)
Δ = 2401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2401}=49$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-49}{2*1}=\frac{-50}{2}
=-25
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+49}{2*1}=\frac{48}{2}
=24
$