2x^2+10=50

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Solution for 2x^2+10=50 equation:



2x^2+10=50
We move all terms to the left:
2x^2+10-(50)=0
We add all the numbers together, and all the variables
2x^2-40=0
a = 2; b = 0; c = -40;
Δ = b2-4ac
Δ = 02-4·2·(-40)
Δ = 320
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:
$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{5}}{2*2}=\frac{0-8\sqrt{5}}{4} =-\frac{8\sqrt{5}}{4} =-2\sqrt{5} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{5}}{2*2}=\frac{0+8\sqrt{5}}{4} =\frac{8\sqrt{5}}{4} =2\sqrt{5} $

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