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3x^2+39x+112=0
a = 3; b = 39; c = +112;
Δ = b2-4ac
Δ = 392-4·3·112
Δ = 177
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{-(39)-\sqrt{177}}{2*3}=\frac{-39-\sqrt{177}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+\sqrt{177}}{2*3}=\frac{-39+\sqrt{177}}{6} $
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