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0.05x^2+55x-11250=0
a = 0.05; b = 55; c = -11250;
Δ = b2-4ac
Δ = 552-4·0.05·(-11250)
Δ = 5275
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5275}=\sqrt{25*211}=\sqrt{25}*\sqrt{211}=5\sqrt{211}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(55)-5\sqrt{211}}{2*0.05}=\frac{-55-5\sqrt{211}}{0.1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(55)+5\sqrt{211}}{2*0.05}=\frac{-55+5\sqrt{211}}{0.1} $
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