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x^2+3x-4096=0
a = 1; b = 3; c = -4096;
Δ = b2-4ac
Δ = 32-4·1·(-4096)
Δ = 16393
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{16393}=\sqrt{169*97}=\sqrt{169}*\sqrt{97}=13\sqrt{97}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-13\sqrt{97}}{2*1}=\frac{-3-13\sqrt{97}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+13\sqrt{97}}{2*1}=\frac{-3+13\sqrt{97}}{2} $
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