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4u^2-9u+5=0
a = 4; b = -9; c = +5;
Δ = b2-4ac
Δ = -92-4·4·5
Δ = 1
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{1}=1$$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-1}{2*4}=\frac{8}{8} =1 $$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+1}{2*4}=\frac{10}{8} =1+1/4 $
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