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16-2x^2=0
a = -2; b = 0; c = +16;
Δ = b2-4ac
Δ = 02-4·(-2)·16
Δ = 128
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{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{2}}{2*-2}=\frac{0-8\sqrt{2}}{-4} =-\frac{8\sqrt{2}}{-4} =-\frac{2\sqrt{2}}{-1}x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{2}}{2*-2}=\frac{0+8\sqrt{2}}{-4} =\frac{8\sqrt{2}}{-4} =\frac{2\sqrt{2}}{-1}
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