I(X)=240x+0.05x2

Solution for I(X)=240x+0.05x2 equation:

(I)=240I+0.05I^2

We move all terms to the left:

(I)-(240I+0.05I^2)=0

We get rid of parentheses

-0.05I^2-240I+I=0

We add all the numbers together, and all the variables

-0.05I^2-239I=0

a = -0.05; b = -239; c = 0;
Δ = b2-4ac
Δ = -2392-4·(-0.05)·0
Δ = 57121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$I_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$I_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{57121}=239$
$I_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-239)-239}{2*-0.05}=\frac{0}{-0.1}
=0
$
$I_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-239)+239}{2*-0.05}=\frac{478}{-0.1}
=-4780
$