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5x^2-19x+17=0
a = 5; b = -19; c = +17;
Δ = b2-4ac
Δ = -192-4·5·17
Δ = 21
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-\sqrt{21}}{2*5}=\frac{19-\sqrt{21}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{21}}{2*5}=\frac{19+\sqrt{21}}{10} $
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