0.02x+0.03x(11,000-x)=290

Solution for 0.02x+0.03x(11,000-x)=290 equation:

0.02x+0.03x(11.000-x)=290

We move all terms to the left:

0.02x+0.03x(11.000-x)-(290)=0

We add all the numbers together, and all the variables

0.02x+0.03x(-1x+11)-290=0

We multiply parentheses

0x^2+0.02x+0x-290=0

We add all the numbers together, and all the variables

x^2+1.02x-290=0

a = 1; b = 1.02; c = -290;
Δ = b2-4ac
Δ = 1.022-4·1·(-290)
Δ = 1161.0404
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{-(1.02)-\sqrt{1161.0404}}{2*1}=\frac{-1.02-\sqrt{1161.0404}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.02)+\sqrt{1161.0404}}{2*1}=\frac{-1.02+\sqrt{1161.0404}}{2}
$