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x^2-41=5x-5
We move all terms to the left:
x^2-41-(5x-5)=0
We get rid of parentheses
x^2-5x+5-41=0
We add all the numbers together, and all the variables
x^2-5x-36=0
a = 1; b = -5; c = -36;
Δ = b2-4ac
Δ = -52-4·1·(-36)
Δ = 169
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{169}=13$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-13}{2*1}=\frac{-8}{2} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+13}{2*1}=\frac{18}{2} =9 $
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