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6x^2-3=11x
We move all terms to the left:
6x^2-3-(11x)=0
a = 6; b = -11; c = -3;
Δ = b2-4ac
Δ = -112-4·6·(-3)
Δ = 193
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{193}}{2*6}=\frac{11-\sqrt{193}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{193}}{2*6}=\frac{11+\sqrt{193}}{12} $
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