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3x^2-120x+800=0
a = 3; b = -120; c = +800;
Δ = b2-4ac
Δ = -1202-4·3·800
Δ = 4800
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4800}=\sqrt{1600*3}=\sqrt{1600}*\sqrt{3}=40\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-120)-40\sqrt{3}}{2*3}=\frac{120-40\sqrt{3}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-120)+40\sqrt{3}}{2*3}=\frac{120+40\sqrt{3}}{6} $
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