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0.4x^2+1x+0.4=0
We add all the numbers together, and all the variables
0.4x^2+x+0.4=0
a = 0.4; b = 1; c = +0.4;
Δ = b2-4ac
Δ = 12-4·0.4·0.4
Δ = 0.36
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{-(1)-\sqrt{0.36}}{2*0.4}=\frac{-1-\sqrt{0.36}}{0.8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{0.36}}{2*0.4}=\frac{-1+\sqrt{0.36}}{0.8} $
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