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14x^2+5x-3=0
a = 14; b = 5; c = -3;
Δ = b2-4ac
Δ = 52-4·14·(-3)
Δ = 193
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{-(5)-\sqrt{193}}{2*14}=\frac{-5-\sqrt{193}}{28} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{193}}{2*14}=\frac{-5+\sqrt{193}}{28} $
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