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3x^2+14x+11=0
a = 3; b = 14; c = +11;
Δ = b2-4ac
Δ = 142-4·3·11
Δ = 64
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{64}=8$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-8}{2*3}=\frac{-22}{6} =-3+2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+8}{2*3}=\frac{-6}{6} =-1 $
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