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0.5x^2-4x-10=0
a = 0.5; b = -4; c = -10;
Δ = b2-4ac
Δ = -42-4·0.5·(-10)
Δ = 36
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{36}=6$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-6}{2*0.5}=\frac{-2}{1} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+6}{2*0.5}=\frac{10}{1} =10 $
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