4r(r+3)-3r=19

Solution for 4r(r+3)-3r=19 equation:

4r(r+3)-3r=19

We move all terms to the left:

4r(r+3)-3r-(19)=0

We add all the numbers together, and all the variables

-3r+4r(r+3)-19=0

We multiply parentheses

4r^2-3r+12r-19=0

We add all the numbers together, and all the variables

4r^2+9r-19=0

a = 4; b = 9; c = -19;
Δ = b2-4ac
Δ = 92-4·4·(-19)
Δ = 385
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{-(9)-\sqrt{385}}{2*4}=\frac{-9-\sqrt{385}}{8}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{385}}{2*4}=\frac{-9+\sqrt{385}}{8}
$