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3n^2+5n-144=0
a = 3; b = 5; c = -144;
Δ = b2-4ac
Δ = 52-4·3·(-144)
Δ = 1753
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{1753}}{2*3}=\frac{-5-\sqrt{1753}}{6} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{1753}}{2*3}=\frac{-5+\sqrt{1753}}{6} $
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