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14x^2+x-105=0
a = 14; b = 1; c = -105;
Δ = b2-4ac
Δ = 12-4·14·(-105)
Δ = 5881
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{-(1)-\sqrt{5881}}{2*14}=\frac{-1-\sqrt{5881}}{28} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{5881}}{2*14}=\frac{-1+\sqrt{5881}}{28} $
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