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3x^2-80x+160=0
a = 3; b = -80; c = +160;
Δ = b2-4ac
Δ = -802-4·3·160
Δ = 4480
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4480}=\sqrt{64*70}=\sqrt{64}*\sqrt{70}=8\sqrt{70}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-8\sqrt{70}}{2*3}=\frac{80-8\sqrt{70}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+8\sqrt{70}}{2*3}=\frac{80+8\sqrt{70}}{6} $
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