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14x^2+19x-3=0
a = 14; b = 19; c = -3;
Δ = b2-4ac
Δ = 192-4·14·(-3)
Δ = 529
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{529}=23$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-23}{2*14}=\frac{-42}{28} =-1+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+23}{2*14}=\frac{4}{28} =1/7 $
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