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6x^2+3x-6=0
a = 6; b = 3; c = -6;
Δ = b2-4ac
Δ = 32-4·6·(-6)
Δ = 153
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{153}=\sqrt{9*17}=\sqrt{9}*\sqrt{17}=3\sqrt{17}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{17}}{2*6}=\frac{-3-3\sqrt{17}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{17}}{2*6}=\frac{-3+3\sqrt{17}}{12} $
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