R(x)=-0.14x^2+420x

Solution for R(x)=-0.14x^2+420x equation:

(R)=-0.14R^2+420R

We move all terms to the left:

(R)-(-0.14R^2+420R)=0

We get rid of parentheses

0.14R^2-420R+R=0

We add all the numbers together, and all the variables

0.14R^2-419R=0

a = 0.14; b = -419; c = 0;
Δ = b2-4ac
Δ = -4192-4·0.14·0
Δ = 175561
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{175561}=419$
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-419)-419}{2*0.14}=\frac{0}{0.28}
=0
$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-419)+419}{2*0.14}=\frac{838}{0.28}
=2992+0.2399999999999/0.28
$