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14x^2-5x-1=0
a = 14; b = -5; c = -1;
Δ = b2-4ac
Δ = -52-4·14·(-1)
Δ = 81
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{81}=9$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-9}{2*14}=\frac{-4}{28} =-1/7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+9}{2*14}=\frac{14}{28} =1/2 $
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