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x^2=8x+16=6x
We move all terms to the left:
x^2-(8x+16)=0
We get rid of parentheses
x^2-8x-16=0
a = 1; b = -8; c = -16;
Δ = b2-4ac
Δ = -82-4·1·(-16)
Δ = 128
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8\sqrt{2}}{2*1}=\frac{8-8\sqrt{2}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8\sqrt{2}}{2*1}=\frac{8+8\sqrt{2}}{2} $
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