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u^2-12u-13=0
a = 1; b = -12; c = -13;
Δ = b2-4ac
Δ = -122-4·1·(-13)
Δ = 196
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{196}=14$$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-14}{2*1}=\frac{-2}{2} =-1 $$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+14}{2*1}=\frac{26}{2} =13 $
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