10-18x+x^2=0

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Solution for 10-18x+x^2=0 equation:



10-18x+x^2=0
a = 1; b = -18; c = +10;
Δ = b2-4ac
Δ = -182-4·1·10
Δ = 284
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:
$\sqrt{\Delta}=\sqrt{284}=\sqrt{4*71}=\sqrt{4}*\sqrt{71}=2\sqrt{71}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{71}}{2*1}=\frac{18-2\sqrt{71}}{2} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{71}}{2*1}=\frac{18+2\sqrt{71}}{2} $

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