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28x^2+80x-168=0
a = 28; b = 80; c = -168;
Δ = b2-4ac
Δ = 802-4·28·(-168)
Δ = 25216
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{25216}=\sqrt{64*394}=\sqrt{64}*\sqrt{394}=8\sqrt{394}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-8\sqrt{394}}{2*28}=\frac{-80-8\sqrt{394}}{56} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+8\sqrt{394}}{2*28}=\frac{-80+8\sqrt{394}}{56} $
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