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3r^2-2r-8=0
a = 3; b = -2; c = -8;
Δ = b2-4ac
Δ = -22-4·3·(-8)
Δ = 100
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{100}=10$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-10}{2*3}=\frac{-8}{6} =-1+1/3 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+10}{2*3}=\frac{12}{6} =2 $
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