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2u^2-29u+25=0
a = 2; b = -29; c = +25;
Δ = b2-4ac
Δ = -292-4·2·25
Δ = 641
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}$$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-\sqrt{641}}{2*2}=\frac{29-\sqrt{641}}{4} $$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+\sqrt{641}}{2*2}=\frac{29+\sqrt{641}}{4} $
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