4r^2=r+19

Solution for 4r^2=r+19 equation:

4r^2=r+19

We move all terms to the left:

4r^2-(r+19)=0

We get rid of parentheses

4r^2-r-19=0

We add all the numbers together, and all the variables

4r^2-1r-19=0

a = 4; b = -1; c = -19;
Δ = b2-4ac
Δ = -12-4·4·(-19)
Δ = 305
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{305}}{2*4}=\frac{1-\sqrt{305}}{8}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{305}}{2*4}=\frac{1+\sqrt{305}}{8}
$