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3x^2+120x+32=0
a = 3; b = 120; c = +32;
Δ = b2-4ac
Δ = 1202-4·3·32
Δ = 14016
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{14016}=\sqrt{64*219}=\sqrt{64}*\sqrt{219}=8\sqrt{219}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(120)-8\sqrt{219}}{2*3}=\frac{-120-8\sqrt{219}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(120)+8\sqrt{219}}{2*3}=\frac{-120+8\sqrt{219}}{6} $
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