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x^2+8x+16=26
We move all terms to the left:
x^2+8x+16-(26)=0
We add all the numbers together, and all the variables
x^2+8x-10=0
a = 1; b = 8; c = -10;
Δ = b2-4ac
Δ = 82-4·1·(-10)
Δ = 104
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{104}=\sqrt{4*26}=\sqrt{4}*\sqrt{26}=2\sqrt{26}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{26}}{2*1}=\frac{-8-2\sqrt{26}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{26}}{2*1}=\frac{-8+2\sqrt{26}}{2} $
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