3r2-19r=14

Solution for 3r2-19r=14 equation:

3r^2-19r=14

We move all terms to the left:

3r^2-19r-(14)=0

a = 3; b = -19; c = -14;
Δ = b2-4ac
Δ = -192-4·3·(-14)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-23}{2*3}=\frac{-4}{6}
=-2/3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+23}{2*3}=\frac{42}{6}
=7
$