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3x^2+0.4x-0.04=0
a = 3; b = 0.4; c = -0.04;
Δ = b2-4ac
Δ = 0.42-4·3·(-0.04)
Δ = 0.64
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{-(0.4)-\sqrt{0.64}}{2*3}=\frac{-0.4-\sqrt{0.64}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.4)+\sqrt{0.64}}{2*3}=\frac{-0.4+\sqrt{0.64}}{6} $
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