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2x^2=20.5^2
We move all terms to the left:
2x^2-(20.5^2)=0
We add all the numbers together, and all the variables
2x^2-420.25=0
a = 2; b = 0; c = -420.25;
Δ = b2-4ac
Δ = 02-4·2·(-420.25)
Δ = 3362
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3362}=\sqrt{1681*2}=\sqrt{1681}*\sqrt{2}=41\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-41\sqrt{2}}{2*2}=\frac{0-41\sqrt{2}}{4} =-\frac{41\sqrt{2}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+41\sqrt{2}}{2*2}=\frac{0+41\sqrt{2}}{4} =\frac{41\sqrt{2}}{4} $
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