500+1300x-100x^2=4100

Solution for 500+1300x-100x^2=4100 equation:

500+1300x-100x^2=4100

We move all terms to the left:

500+1300x-100x^2-(4100)=0

We add all the numbers together, and all the variables

-100x^2+1300x-3600=0

a = -100; b = 1300; c = -3600;
Δ = b2-4ac
Δ = 13002-4·(-100)·(-3600)
Δ = 250000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{250000}=500$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1300)-500}{2*-100}=\frac{-1800}{-200}
=+9
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1300)+500}{2*-100}=\frac{-800}{-200}
=+4
$