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3x^2-49x-144=0
a = 3; b = -49; c = -144;
Δ = b2-4ac
Δ = -492-4·3·(-144)
Δ = 4129
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{-(-49)-\sqrt{4129}}{2*3}=\frac{49-\sqrt{4129}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-49)+\sqrt{4129}}{2*3}=\frac{49+\sqrt{4129}}{6} $
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