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15a^2-28a+5=0
a = 15; b = -28; c = +5;
Δ = b2-4ac
Δ = -282-4·15·5
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{484}=22$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-22}{2*15}=\frac{6}{30} =1/5 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+22}{2*15}=\frac{50}{30} =1+2/3 $
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