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6x^2+12x-11=0
a = 6; b = 12; c = -11;
Δ = b2-4ac
Δ = 122-4·6·(-11)
Δ = 408
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{408}=\sqrt{4*102}=\sqrt{4}*\sqrt{102}=2\sqrt{102}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{102}}{2*6}=\frac{-12-2\sqrt{102}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{102}}{2*6}=\frac{-12+2\sqrt{102}}{12} $
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