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3x^2-240x+20=0
a = 3; b = -240; c = +20;
Δ = b2-4ac
Δ = -2402-4·3·20
Δ = 57360
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{57360}=\sqrt{16*3585}=\sqrt{16}*\sqrt{3585}=4\sqrt{3585}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-240)-4\sqrt{3585}}{2*3}=\frac{240-4\sqrt{3585}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-240)+4\sqrt{3585}}{2*3}=\frac{240+4\sqrt{3585}}{6} $
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