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x^2+18x=326
We move all terms to the left:
x^2+18x-(326)=0
a = 1; b = 18; c = -326;
Δ = b2-4ac
Δ = 182-4·1·(-326)
Δ = 1628
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1628}=\sqrt{4*407}=\sqrt{4}*\sqrt{407}=2\sqrt{407}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{407}}{2*1}=\frac{-18-2\sqrt{407}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{407}}{2*1}=\frac{-18+2\sqrt{407}}{2} $
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