125x=3100+.1x^2

Solution for 125x=3100+.1x^2 equation:

125x=3100+.1x^2

We move all terms to the left:

125x-(3100+.1x^2)=0

We get rid of parentheses

-.1x^2+125x-3100=0

We add all the numbers together, and all the variables

-0.1x^2+125x-3100=0

a = -0.1; b = 125; c = -3100;
Δ = b2-4ac
Δ = 1252-4·(-0.1)·(-3100)
Δ = 14385
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{-(125)-\sqrt{14385}}{2*-0.1}=\frac{-125-\sqrt{14385}}{-0.2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(125)+\sqrt{14385}}{2*-0.1}=\frac{-125+\sqrt{14385}}{-0.2}
$