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16u^2-41u+25=0
a = 16; b = -41; c = +25;
Δ = b2-4ac
Δ = -412-4·16·25
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{81}=9$$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-41)-9}{2*16}=\frac{32}{32} =1 $$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+9}{2*16}=\frac{50}{32} =1+9/16 $
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