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x^2+180x-4800=0
a = 1; b = 180; c = -4800;
Δ = b2-4ac
Δ = 1802-4·1·(-4800)
Δ = 51600
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{51600}=\sqrt{400*129}=\sqrt{400}*\sqrt{129}=20\sqrt{129}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(180)-20\sqrt{129}}{2*1}=\frac{-180-20\sqrt{129}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(180)+20\sqrt{129}}{2*1}=\frac{-180+20\sqrt{129}}{2} $
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