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10r^2-17r-6=0
a = 10; b = -17; c = -6;
Δ = b2-4ac
Δ = -172-4·10·(-6)
Δ = 529
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{529}=23$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-23}{2*10}=\frac{-6}{20} =-3/10 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+23}{2*10}=\frac{40}{20} =2 $
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