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325x^2+200x-125=0
a = 325; b = 200; c = -125;
Δ = b2-4ac
Δ = 2002-4·325·(-125)
Δ = 202500
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{202500}=450$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(200)-450}{2*325}=\frac{-650}{650} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(200)+450}{2*325}=\frac{250}{650} =5/13 $
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