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0.04x^2+5x+140=0
a = 0.04; b = 5; c = +140;
Δ = b2-4ac
Δ = 52-4·0.04·140
Δ = 2.6
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{-(5)-\sqrt{2.6}}{2*0.04}=\frac{-5-\sqrt{2.6}}{0.08} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{2.6}}{2*0.04}=\frac{-5+\sqrt{2.6}}{0.08} $
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