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-x^2+2x-0.5=0
We add all the numbers together, and all the variables
-1x^2+2x-0.5=0
a = -1; b = 2; c = -0.5;
Δ = b2-4ac
Δ = 22-4·(-1)·(-0.5)
Δ = 2
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{-(2)-\sqrt{2}}{2*-1}=\frac{-2-\sqrt{2}}{-2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+\sqrt{2}}{2*-1}=\frac{-2+\sqrt{2}}{-2} $
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