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3x^2-80x+400=0
a = 3; b = -80; c = +400;
Δ = b2-4ac
Δ = -802-4·3·400
Δ = 1600
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1600}=40$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-40}{2*3}=\frac{40}{6} =6+2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+40}{2*3}=\frac{120}{6} =20 $
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