3r^2+6r-189=0

Solution for 3r^2+6r-189=0 equation:

3r^2+6r-189=0

a = 3; b = 6; c = -189;
Δ = b2-4ac
Δ = 62-4·3·(-189)
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-48}{2*3}=\frac{-54}{6}
=-9
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+48}{2*3}=\frac{42}{6}
=7
$