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128=2x^2
We move all terms to the left:
128-(2x^2)=0
a = -2; b = 0; c = +128;
Δ = b2-4ac
Δ = 02-4·(-2)·128
Δ = 1024
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1024}=32$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32}{2*-2}=\frac{-32}{-4} =+8 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32}{2*-2}=\frac{32}{-4} =-8 $
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