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4x^2-240x+1200=0
a = 4; b = -240; c = +1200;
Δ = b2-4ac
Δ = -2402-4·4·1200
Δ = 38400
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{38400}=\sqrt{6400*6}=\sqrt{6400}*\sqrt{6}=80\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-240)-80\sqrt{6}}{2*4}=\frac{240-80\sqrt{6}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-240)+80\sqrt{6}}{2*4}=\frac{240+80\sqrt{6}}{8} $
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