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15x^2+x-14=0
a = 15; b = 1; c = -14;
Δ = b2-4ac
Δ = 12-4·15·(-14)
Δ = 841
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{841}=29$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-29}{2*15}=\frac{-30}{30} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+29}{2*15}=\frac{28}{30} =14/15 $
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