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3x^2+11x-4=0
a = 3; b = 11; c = -4;
Δ = b2-4ac
Δ = 112-4·3·(-4)
Δ = 169
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{169}=13$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-13}{2*3}=\frac{-24}{6} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+13}{2*3}=\frac{2}{6} =1/3 $
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