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2x^2+2x-1.25=0
a = 2; b = 2; c = -1.25;
Δ = b2-4ac
Δ = 22-4·2·(-1.25)
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{-(2)-\sqrt{14}}{2*2}=\frac{-2-\sqrt{14}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+\sqrt{14}}{2*2}=\frac{-2+\sqrt{14}}{4} $
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