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3x^2+44x-120=0
a = 3; b = 44; c = -120;
Δ = b2-4ac
Δ = 442-4·3·(-120)
Δ = 3376
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3376}=\sqrt{16*211}=\sqrt{16}*\sqrt{211}=4\sqrt{211}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(44)-4\sqrt{211}}{2*3}=\frac{-44-4\sqrt{211}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(44)+4\sqrt{211}}{2*3}=\frac{-44+4\sqrt{211}}{6} $
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