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x2+13=8x
We move all terms to the left:
x2+13-(8x)=0
We add all the numbers together, and all the variables
x^2-8x+13=0
a = 1; b = -8; c = +13;
Δ = b2-4ac
Δ = -82-4·1·13
Δ = 12
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{3}}{2*1}=\frac{8-2\sqrt{3}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{3}}{2*1}=\frac{8+2\sqrt{3}}{2} $
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