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3x^2+27x-61=0
a = 3; b = 27; c = -61;
Δ = b2-4ac
Δ = 272-4·3·(-61)
Δ = 1461
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{-(27)-\sqrt{1461}}{2*3}=\frac{-27-\sqrt{1461}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+\sqrt{1461}}{2*3}=\frac{-27+\sqrt{1461}}{6} $
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