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6x^2+12x-92=0
a = 6; b = 12; c = -92;
Δ = b2-4ac
Δ = 122-4·6·(-92)
Δ = 2352
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2352}=\sqrt{784*3}=\sqrt{784}*\sqrt{3}=28\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-28\sqrt{3}}{2*6}=\frac{-12-28\sqrt{3}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+28\sqrt{3}}{2*6}=\frac{-12+28\sqrt{3}}{12} $
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