x2+(10-2x)2=20

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Solution for x2+(10-2x)2=20 equation:



x2+(10-2x)2=20
We move all terms to the left:
x2+(10-2x)2-(20)=0
We add all the numbers together, and all the variables
x2+(-2x+10)2-20=0
We add all the numbers together, and all the variables
x^2+(-2x+10)2-20=0
We multiply parentheses
x^2-4x+20-20=0
We add all the numbers together, and all the variables
x^2-4x=0
a = 1; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·1·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*1}=\frac{0}{2} =0 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*1}=\frac{8}{2} =4 $

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