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3x^2-18x-325=0
a = 3; b = -18; c = -325;
Δ = b2-4ac
Δ = -182-4·3·(-325)
Δ = 4224
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4224}=\sqrt{64*66}=\sqrt{64}*\sqrt{66}=8\sqrt{66}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-8\sqrt{66}}{2*3}=\frac{18-8\sqrt{66}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+8\sqrt{66}}{2*3}=\frac{18+8\sqrt{66}}{6} $
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