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15r^2-17r+4=0
a = 15; b = -17; c = +4;
Δ = b2-4ac
Δ = -172-4·15·4
Δ = 49
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{49}=7$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-7}{2*15}=\frac{10}{30} =1/3 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+7}{2*15}=\frac{24}{30} =4/5 $
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