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x^2-18x+16=25
We move all terms to the left:
x^2-18x+16-(25)=0
We add all the numbers together, and all the variables
x^2-18x-9=0
a = 1; b = -18; c = -9;
Δ = b2-4ac
Δ = -182-4·1·(-9)
Δ = 360
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{360}=\sqrt{36*10}=\sqrt{36}*\sqrt{10}=6\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-6\sqrt{10}}{2*1}=\frac{18-6\sqrt{10}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+6\sqrt{10}}{2*1}=\frac{18+6\sqrt{10}}{2} $
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