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14x^2+18x-1=0
a = 14; b = 18; c = -1;
Δ = b2-4ac
Δ = 182-4·14·(-1)
Δ = 380
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{380}=\sqrt{4*95}=\sqrt{4}*\sqrt{95}=2\sqrt{95}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{95}}{2*14}=\frac{-18-2\sqrt{95}}{28} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{95}}{2*14}=\frac{-18+2\sqrt{95}}{28} $
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