6r^2+17r+12=0

Solution for 6r^2+17r+12=0 equation:

6r^2+17r+12=0

a = 6; b = 17; c = +12;
Δ = b2-4ac
Δ = 172-4·6·12
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-1}{2*6}=\frac{-18}{12}
=-1+1/2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+1}{2*6}=\frac{-16}{12}
=-1+1/3
$