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0.05x^2-4x+80.8=40
We move all terms to the left:
0.05x^2-4x+80.8-(40)=0
We add all the numbers together, and all the variables
0.05x^2-4x+40.8=0
a = 0.05; b = -4; c = +40.8;
Δ = b2-4ac
Δ = -42-4·0.05·40.8
Δ = 7.84
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{-(-4)-\sqrt{7.84}}{2*0.05}=\frac{4-\sqrt{7.84}}{0.1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+\sqrt{7.84}}{2*0.05}=\frac{4+\sqrt{7.84}}{0.1} $
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