C(x)=0.001x^2+0.7x+10,R(x)=3x

Solution for C(x)=0.001x^2+0.7x+10,R(x)=3x equation:

(C)=0.001C^2+0.7C+10.(C)=3C

We move all terms to the left:

(C)-(0.001C^2+0.7C+10.(C))=0

We get rid of parentheses

-0.001C^2+C-0.7C-10.C=0

We add all the numbers together, and all the variables

-0.001C^2-9.7C=0

a = -0.001; b = -9.7; c = 0;
Δ = b2-4ac
Δ = -9.72-4·(-0.001)·0
Δ = 94.09
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$C_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$C_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$C_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9.7)-\sqrt{94.09}}{2*-0.001}=\frac{9.7-\sqrt{94.09}}{-0.002}
$
$C_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9.7)+\sqrt{94.09}}{2*-0.001}=\frac{9.7+\sqrt{94.09}}{-0.002}
$