3r^2+6r-105=0

Solution for 3r^2+6r-105=0 equation:

3r^2+6r-105=0

a = 3; b = 6; c = -105;
Δ = b2-4ac
Δ = 62-4·3·(-105)
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-36}{2*3}=\frac{-42}{6}
=-7
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+36}{2*3}=\frac{30}{6}
=5
$