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5u^2-28u+15=0
a = 5; b = -28; c = +15;
Δ = b2-4ac
Δ = -282-4·5·15
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{484}=22$$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-22}{2*5}=\frac{6}{10} =3/5 $$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+22}{2*5}=\frac{50}{10} =5 $
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