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3x^2+23x-34=0
a = 3; b = 23; c = -34;
Δ = b2-4ac
Δ = 232-4·3·(-34)
Δ = 937
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-\sqrt{937}}{2*3}=\frac{-23-\sqrt{937}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+\sqrt{937}}{2*3}=\frac{-23+\sqrt{937}}{6} $
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