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96=2u^2
We move all terms to the left:
96-(2u^2)=0
a = -2; b = 0; c = +96;
Δ = b2-4ac
Δ = 02-4·(-2)·96
Δ = 768
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{768}=\sqrt{256*3}=\sqrt{256}*\sqrt{3}=16\sqrt{3}$$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{3}}{2*-2}=\frac{0-16\sqrt{3}}{-4} =-\frac{16\sqrt{3}}{-4} =-\frac{4\sqrt{3}}{-1} $$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{3}}{2*-2}=\frac{0+16\sqrt{3}}{-4} =\frac{16\sqrt{3}}{-4} =\frac{4\sqrt{3}}{-1} $
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