x=-4x^2+320x

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Solution for x=-4x^2+320x equation:



x=-4x^2+320x
We move all terms to the left:
x-(-4x^2+320x)=0
We get rid of parentheses
4x^2-320x+x=0
We add all the numbers together, and all the variables
4x^2-319x=0
a = 4; b = -319; c = 0;
Δ = b2-4ac
Δ = -3192-4·4·0
Δ = 101761
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{101761}=319$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-319)-319}{2*4}=\frac{0}{8} =0 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-319)+319}{2*4}=\frac{638}{8} =79+3/4 $

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