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2x^2+11x-189=0
a = 2; b = 11; c = -189;
Δ = b2-4ac
Δ = 112-4·2·(-189)
Δ = 1633
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{-(11)-\sqrt{1633}}{2*2}=\frac{-11-\sqrt{1633}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{1633}}{2*2}=\frac{-11+\sqrt{1633}}{4} $
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