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11x^2-33x+12=0
a = 11; b = -33; c = +12;
Δ = b2-4ac
Δ = -332-4·11·12
Δ = 561
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{-(-33)-\sqrt{561}}{2*11}=\frac{33-\sqrt{561}}{22} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-33)+\sqrt{561}}{2*11}=\frac{33+\sqrt{561}}{22} $
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