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14x^2+x-3=0
a = 14; b = 1; c = -3;
Δ = b2-4ac
Δ = 12-4·14·(-3)
Δ = 169
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{169}=13$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-13}{2*14}=\frac{-14}{28} =-1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+13}{2*14}=\frac{12}{28} =3/7 $
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