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3x^2+19x-40=0
a = 3; b = 19; c = -40;
Δ = b2-4ac
Δ = 192-4·3·(-40)
Δ = 841
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{841}=29$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-29}{2*3}=\frac{-48}{6} =-8 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+29}{2*3}=\frac{10}{6} =1+2/3 $
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