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x^2-18x=-16
We move all terms to the left:
x^2-18x-(-16)=0
We add all the numbers together, and all the variables
x^2-18x+16=0
a = 1; b = -18; c = +16;
Δ = b2-4ac
Δ = -182-4·1·16
Δ = 260
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{260}=\sqrt{4*65}=\sqrt{4}*\sqrt{65}=2\sqrt{65}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{65}}{2*1}=\frac{18-2\sqrt{65}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{65}}{2*1}=\frac{18+2\sqrt{65}}{2} $
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