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0.1x^2+104x-3100=0
a = 0.1; b = 104; c = -3100;
Δ = b2-4ac
Δ = 1042-4·0.1·(-3100)
Δ = 12056
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{12056}=\sqrt{4*3014}=\sqrt{4}*\sqrt{3014}=2\sqrt{3014}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(104)-2\sqrt{3014}}{2*0.1}=\frac{-104-2\sqrt{3014}}{0.2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(104)+2\sqrt{3014}}{2*0.1}=\frac{-104+2\sqrt{3014}}{0.2} $
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