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14x^2+3x-11=0
a = 14; b = 3; c = -11;
Δ = b2-4ac
Δ = 32-4·14·(-11)
Δ = 625
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{625}=25$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-25}{2*14}=\frac{-28}{28} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+25}{2*14}=\frac{22}{28} =11/14 $
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