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6r^2-5r-6=0
a = 6; b = -5; c = -6;
Δ = b2-4ac
Δ = -52-4·6·(-6)
Δ = 169
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{169}=13$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-13}{2*6}=\frac{-8}{12} =-2/3 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+13}{2*6}=\frac{18}{12} =1+1/2 $
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