R(x)=4x^2+9x+3

Solution for R(x)=4x^2+9x+3 equation:

(R)=4R^2+9R+3

We move all terms to the left:

(R)-(4R^2+9R+3)=0

We get rid of parentheses

-4R^2+R-9R-3=0

We add all the numbers together, and all the variables

-4R^2-8R-3=0

a = -4; b = -8; c = -3;
Δ = b2-4ac
Δ = -82-4·(-4)·(-3)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4}{2*-4}=\frac{4}{-8}
=-1/2
$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4}{2*-4}=\frac{12}{-8}
=-1+1/2
$