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x^2-3600-3x=0
a = 1; b = -3; c = -3600;
Δ = b2-4ac
Δ = -32-4·1·(-3600)
Δ = 14409
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{14409}=\sqrt{9*1601}=\sqrt{9}*\sqrt{1601}=3\sqrt{1601}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3\sqrt{1601}}{2*1}=\frac{3-3\sqrt{1601}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3\sqrt{1601}}{2*1}=\frac{3+3\sqrt{1601}}{2} $
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