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a^2-26a+69=0
a = 1; b = -26; c = +69;
Δ = b2-4ac
Δ = -262-4·1·69
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{400}=20$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-20}{2*1}=\frac{6}{2} =3 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+20}{2*1}=\frac{46}{2} =23 $
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