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2x^2-16x-7396=0
a = 2; b = -16; c = -7396;
Δ = b2-4ac
Δ = -162-4·2·(-7396)
Δ = 59424
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{59424}=\sqrt{16*3714}=\sqrt{16}*\sqrt{3714}=4\sqrt{3714}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{3714}}{2*2}=\frac{16-4\sqrt{3714}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{3714}}{2*2}=\frac{16+4\sqrt{3714}}{4} $
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