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3x^2-44x+120=0
a = 3; b = -44; c = +120;
Δ = b2-4ac
Δ = -442-4·3·120
Δ = 496
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{496}=\sqrt{16*31}=\sqrt{16}*\sqrt{31}=4\sqrt{31}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-44)-4\sqrt{31}}{2*3}=\frac{44-4\sqrt{31}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-44)+4\sqrt{31}}{2*3}=\frac{44+4\sqrt{31}}{6} $
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