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2x^2+200x-3200=0
a = 2; b = 200; c = -3200;
Δ = b2-4ac
Δ = 2002-4·2·(-3200)
Δ = 65600
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{65600}=\sqrt{1600*41}=\sqrt{1600}*\sqrt{41}=40\sqrt{41}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(200)-40\sqrt{41}}{2*2}=\frac{-200-40\sqrt{41}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(200)+40\sqrt{41}}{2*2}=\frac{-200+40\sqrt{41}}{4} $
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