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24x^2-128=0
a = 24; b = 0; c = -128;
Δ = b2-4ac
Δ = 02-4·24·(-128)
Δ = 12288
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{12288}=\sqrt{4096*3}=\sqrt{4096}*\sqrt{3}=64\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-64\sqrt{3}}{2*24}=\frac{0-64\sqrt{3}}{48} =-\frac{64\sqrt{3}}{48} =-\frac{4\sqrt{3}}{3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+64\sqrt{3}}{2*24}=\frac{0+64\sqrt{3}}{48} =\frac{64\sqrt{3}}{48} =\frac{4\sqrt{3}}{3} $
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