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2x^2-4x-2.5=0
a = 2; b = -4; c = -2.5;
Δ = b2-4ac
Δ = -42-4·2·(-2.5)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-6}{2*2}=\frac{-2}{4} =-1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+6}{2*2}=\frac{10}{4} =2+1/2 $
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