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4u^2+49u+45=0
a = 4; b = 49; c = +45;
Δ = b2-4ac
Δ = 492-4·4·45
Δ = 1681
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{1681}=41$$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(49)-41}{2*4}=\frac{-90}{8} =-11+1/4 $$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(49)+41}{2*4}=\frac{-8}{8} =-1 $
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