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18x^2-16x-18=0
a = 18; b = -16; c = -18;
Δ = b2-4ac
Δ = -162-4·18·(-18)
Δ = 1552
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1552}=\sqrt{16*97}=\sqrt{16}*\sqrt{97}=4\sqrt{97}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{97}}{2*18}=\frac{16-4\sqrt{97}}{36} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{97}}{2*18}=\frac{16+4\sqrt{97}}{36} $
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