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3x^2-3.6x+1=0
a = 3; b = -3.6; c = +1;
Δ = b2-4ac
Δ = -3.62-4·3·1
Δ = 0.96
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{-(-3.6)-\sqrt{0.96}}{2*3}=\frac{3.6-\sqrt{0.96}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3.6)+\sqrt{0.96}}{2*3}=\frac{3.6+\sqrt{0.96}}{6} $
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