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2x^2-3x+0.8=0
a = 2; b = -3; c = +0.8;
Δ = b2-4ac
Δ = -32-4·2·0.8
Δ = 2.6
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{-(-3)-\sqrt{2.6}}{2*2}=\frac{3-\sqrt{2.6}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{2.6}}{2*2}=\frac{3+\sqrt{2.6}}{4} $
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