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20x^2-160=0
a = 20; b = 0; c = -160;
Δ = b2-4ac
Δ = 02-4·20·(-160)
Δ = 12800
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{12800}=\sqrt{6400*2}=\sqrt{6400}*\sqrt{2}=80\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-80\sqrt{2}}{2*20}=\frac{0-80\sqrt{2}}{40} =-\frac{80\sqrt{2}}{40} =-2\sqrt{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+80\sqrt{2}}{2*20}=\frac{0+80\sqrt{2}}{40} =\frac{80\sqrt{2}}{40} =2\sqrt{2} $
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