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-3x^2-6x+11=0
a = -3; b = -6; c = +11;
Δ = b2-4ac
Δ = -62-4·(-3)·11
Δ = 168
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{168}=\sqrt{4*42}=\sqrt{4}*\sqrt{42}=2\sqrt{42}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{42}}{2*-3}=\frac{6-2\sqrt{42}}{-6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{42}}{2*-3}=\frac{6+2\sqrt{42}}{-6} $
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