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3a^2-11a-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:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{169}=13$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-13}{2*3}=\frac{-2}{6} =-1/3 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+13}{2*3}=\frac{24}{6} =4 $
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