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2u^2+25u+12=0
a = 2; b = 25; c = +12;
Δ = b2-4ac
Δ = 252-4·2·12
Δ = 529
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{529}=23$$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-23}{2*2}=\frac{-48}{4} =-12 $$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+23}{2*2}=\frac{-2}{4} =-1/2 $
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