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3x^2+33x+44=0
a = 3; b = 33; c = +44;
Δ = b2-4ac
Δ = 332-4·3·44
Δ = 561
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-\sqrt{561}}{2*3}=\frac{-33-\sqrt{561}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+\sqrt{561}}{2*3}=\frac{-33+\sqrt{561}}{6} $
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