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128-2b^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:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1024}=32$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32}{2*-2}=\frac{-32}{-4} =+8 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32}{2*-2}=\frac{32}{-4} =-8 $
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