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20x^2=40
We move all terms to the left:
20x^2-(40)=0
a = 20; b = 0; c = -40;
Δ = b2-4ac
Δ = 02-4·20·(-40)
Δ = 3200
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3200}=\sqrt{1600*2}=\sqrt{1600}*\sqrt{2}=40\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40\sqrt{2}}{2*20}=\frac{0-40\sqrt{2}}{40} =-\frac{40\sqrt{2}}{40} =-\sqrt{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40\sqrt{2}}{2*20}=\frac{0+40\sqrt{2}}{40} =\frac{40\sqrt{2}}{40} =\sqrt{2} $
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