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-3x^2-23x-40=0
a = -3; b = -23; c = -40;
Δ = b2-4ac
Δ = -232-4·(-3)·(-40)
Δ = 49
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{49}=7$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-7}{2*-3}=\frac{16}{-6} =-2+2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+7}{2*-3}=\frac{30}{-6} =-5 $
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