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0.0498=0.40x-0.4x^2
We move all terms to the left:
0.0498-(0.40x-0.4x^2)=0
We get rid of parentheses
0.4x^2-0.40x+0.0498=0
We add all the numbers together, and all the variables
0.4x^2-0.4x+0.0498=0
a = 0.4; b = -0.4; c = +0.0498;
Δ = b2-4ac
Δ = -0.42-4·0.4·0.0498
Δ = 0.08032
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{0.08032}}{2*0.4}=\frac{0.4-\sqrt{0.08032}}{0.8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.4)+\sqrt{0.08032}}{2*0.4}=\frac{0.4+\sqrt{0.08032}}{0.8} $
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