-.2x^2+80x-2000=0

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Solution for -.2x^2+80x-2000=0 equation:



-.2x^2+80x-2000=0
We add all the numbers together, and all the variables
-0.2x^2+80x-2000=0
a = -0.2; b = 80; c = -2000;
Δ = b2-4ac
Δ = 802-4·(-0.2)·(-2000)
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{-(80)-40\sqrt{3}}{2*-0.2}=\frac{-80-40\sqrt{3}}{-0.4} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+40\sqrt{3}}{2*-0.2}=\frac{-80+40\sqrt{3}}{-0.4} $

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