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3x^2+34x+11=0
a = 3; b = 34; c = +11;
Δ = b2-4ac
Δ = 342-4·3·11
Δ = 1024
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{1024}=32$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-32}{2*3}=\frac{-66}{6} =-11 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+32}{2*3}=\frac{-2}{6} =-1/3 $
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