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3x^2-19x-24=0
a = 3; b = -19; c = -24;
Δ = b2-4ac
Δ = -192-4·3·(-24)
Δ = 649
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{-(-19)-\sqrt{649}}{2*3}=\frac{19-\sqrt{649}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{649}}{2*3}=\frac{19+\sqrt{649}}{6} $
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