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18x^2-6x-7=0
a = 18; b = -6; c = -7;
Δ = b2-4ac
Δ = -62-4·18·(-7)
Δ = 540
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{540}=\sqrt{36*15}=\sqrt{36}*\sqrt{15}=6\sqrt{15}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6\sqrt{15}}{2*18}=\frac{6-6\sqrt{15}}{36} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6\sqrt{15}}{2*18}=\frac{6+6\sqrt{15}}{36} $
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