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x^2+36=8x+33
We move all terms to the left:
x^2+36-(8x+33)=0
We get rid of parentheses
x^2-8x-33+36=0
We add all the numbers together, and all the variables
x^2-8x+3=0
a = 1; b = -8; c = +3;
Δ = b2-4ac
Δ = -82-4·1·3
Δ = 52
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{13}}{2*1}=\frac{8-2\sqrt{13}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{13}}{2*1}=\frac{8+2\sqrt{13}}{2} $
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