2+10x2-5+3=20

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Solution for 2+10x2-5+3=20 equation:



2+10x^2-5+3=20
We move all terms to the left:
2+10x^2-5+3-(20)=0
We add all the numbers together, and all the variables
10x^2-20=0
a = 10; b = 0; c = -20;
Δ = b2-4ac
Δ = 02-4·10·(-20)
Δ = 800
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:
$\sqrt{\Delta}=\sqrt{800}=\sqrt{400*2}=\sqrt{400}*\sqrt{2}=20\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{2}}{2*10}=\frac{0-20\sqrt{2}}{20} =-\frac{20\sqrt{2}}{20} =-\sqrt{2} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{2}}{2*10}=\frac{0+20\sqrt{2}}{20} =\frac{20\sqrt{2}}{20} =\sqrt{2} $

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