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3x^2+0.5x-0.5=0
a = 3; b = 0.5; c = -0.5;
Δ = b2-4ac
Δ = 0.52-4·3·(-0.5)
Δ = 6.25
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{-(0.5)-\sqrt{6.25}}{2*3}=\frac{-0.5-\sqrt{6.25}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.5)+\sqrt{6.25}}{2*3}=\frac{-0.5+\sqrt{6.25}}{6} $
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