150x-(3050+8x+0.1x^2)=500

Solution for 150x-(3050+8x+0.1x^2)=500 equation:

150x-(3050+8x+0.1x^2)=500

We move all terms to the left:

150x-(3050+8x+0.1x^2)-(500)=0

We get rid of parentheses

-0.1x^2-8x+150x-3050-500=0

We add all the numbers together, and all the variables

-0.1x^2+142x-3550=0

a = -0.1; b = 142; c = -3550;
Δ = b2-4ac
Δ = 1422-4·(-0.1)·(-3550)
Δ = 18744
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{18744}=\sqrt{4*4686}=\sqrt{4}*\sqrt{4686}=2\sqrt{4686}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(142)-2\sqrt{4686}}{2*-0.1}=\frac{-142-2\sqrt{4686}}{-0.2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(142)+2\sqrt{4686}}{2*-0.1}=\frac{-142+2\sqrt{4686}}{-0.2}
$