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x^2+1200x=480000
We move all terms to the left:
x^2+1200x-(480000)=0
a = 1; b = 1200; c = -480000;
Δ = b2-4ac
Δ = 12002-4·1·(-480000)
Δ = 3360000
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3360000}=\sqrt{160000*21}=\sqrt{160000}*\sqrt{21}=400\sqrt{21}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1200)-400\sqrt{21}}{2*1}=\frac{-1200-400\sqrt{21}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1200)+400\sqrt{21}}{2*1}=\frac{-1200+400\sqrt{21}}{2} $
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