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5x^2-19x+4=0
a = 5; b = -19; c = +4;
Δ = b2-4ac
Δ = -192-4·5·4
Δ = 281
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{281}}{2*5}=\frac{19-\sqrt{281}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{281}}{2*5}=\frac{19+\sqrt{281}}{10} $
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