R(x)=-2x^2+3x

Solution for R(x)=-2x^2+3x equation:

(R)=-2R^2+3R

We move all terms to the left:

(R)-(-2R^2+3R)=0

We get rid of parentheses

2R^2-3R+R=0

We add all the numbers together, and all the variables

2R^2-2R=0

a = 2; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·2·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*2}=\frac{0}{4}
=0
$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*2}=\frac{4}{4}
=1
$