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3x^2+39x+108=0
a = 3; b = 39; c = +108;
Δ = b2-4ac
Δ = 392-4·3·108
Δ = 225
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{225}=15$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-15}{2*3}=\frac{-54}{6} =-9 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+15}{2*3}=\frac{-24}{6} =-4 $
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