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2x^2-2.5x-0.5=0
a = 2; b = -2.5; c = -0.5;
Δ = b2-4ac
Δ = -2.52-4·2·(-0.5)
Δ = 10.25
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{-(-2.5)-\sqrt{10.25}}{2*2}=\frac{2.5-\sqrt{10.25}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2.5)+\sqrt{10.25}}{2*2}=\frac{2.5+\sqrt{10.25}}{4} $
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