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0.5x^2-0.32=0
a = 0.5; b = 0; c = -0.32;
Δ = b2-4ac
Δ = 02-4·0.5·(-0.32)
Δ = 0.64
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-\sqrt{0.64}}{2*0.5}=\frac{0-\sqrt{0.64}}{1} =-\frac{\sqrt{}}{1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{0.64}}{2*0.5}=\frac{0+\sqrt{0.64}}{1} =\frac{\sqrt{}}{1} $
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