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-3x^2+43x-144=0
a = -3; b = 43; c = -144;
Δ = b2-4ac
Δ = 432-4·(-3)·(-144)
Δ = 121
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{121}=11$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(43)-11}{2*-3}=\frac{-54}{-6} =+9 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(43)+11}{2*-3}=\frac{-32}{-6} =5+1/3 $
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