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3x^2-63x+120=0
a = 3; b = -63; c = +120;
Δ = b2-4ac
Δ = -632-4·3·120
Δ = 2529
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2529}=\sqrt{9*281}=\sqrt{9}*\sqrt{281}=3\sqrt{281}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-63)-3\sqrt{281}}{2*3}=\frac{63-3\sqrt{281}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-63)+3\sqrt{281}}{2*3}=\frac{63+3\sqrt{281}}{6} $
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