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200q-q^2-1000=0
We add all the numbers together, and all the variables
-1q^2+200q-1000=0
a = -1; b = 200; c = -1000;
Δ = b2-4ac
Δ = 2002-4·(-1)·(-1000)
Δ = 36000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{36000}=\sqrt{3600*10}=\sqrt{3600}*\sqrt{10}=60\sqrt{10}$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(200)-60\sqrt{10}}{2*-1}=\frac{-200-60\sqrt{10}}{-2} $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(200)+60\sqrt{10}}{2*-1}=\frac{-200+60\sqrt{10}}{-2} $
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