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4x^2-1280=0
a = 4; b = 0; c = -1280;
Δ = b2-4ac
Δ = 02-4·4·(-1280)
Δ = 20480
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{20480}=\sqrt{4096*5}=\sqrt{4096}*\sqrt{5}=64\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-64\sqrt{5}}{2*4}=\frac{0-64\sqrt{5}}{8} =-\frac{64\sqrt{5}}{8} =-8\sqrt{5} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+64\sqrt{5}}{2*4}=\frac{0+64\sqrt{5}}{8} =\frac{64\sqrt{5}}{8} =8\sqrt{5} $
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