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36-16u-u2=0
We add all the numbers together, and all the variables
-1u^2-16u+36=0
a = -1; b = -16; c = +36;
Δ = b2-4ac
Δ = -162-4·(-1)·36
Δ = 400
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{400}=20$$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-20}{2*-1}=\frac{-4}{-2} =+2 $$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+20}{2*-1}=\frac{36}{-2} =-18 $
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