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-0.5x^2-4x-1=0
a = -0.5; b = -4; c = -1;
Δ = b2-4ac
Δ = -42-4·(-0.5)·(-1)
Δ = 14
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{-(-4)-\sqrt{14}}{2*-0.5}=\frac{4-\sqrt{14}}{-1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+\sqrt{14}}{2*-0.5}=\frac{4+\sqrt{14}}{-1} $
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