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x2+160=26x
We move all terms to the left:
x2+160-(26x)=0
We add all the numbers together, and all the variables
x^2-26x+160=0
a = 1; b = -26; c = +160;
Δ = b2-4ac
Δ = -262-4·1·160
Δ = 36
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{36}=6$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-6}{2*1}=\frac{20}{2} =10 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+6}{2*1}=\frac{32}{2} =16 $
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