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x^2-8x+10=10
We move all terms to the left:
x^2-8x+10-(10)=0
We add all the numbers together, and all the variables
x^2-8x=0
a = 1; b = -8; c = 0;
Δ = b2-4ac
Δ = -82-4·1·0
Δ = 64
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{64}=8$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8}{2*1}=\frac{0}{2} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8}{2*1}=\frac{16}{2} =8 $
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