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(2/3)u=16
We move all terms to the left:
(2/3)u-(16)=0
Domain of the equation: 3)u!=0We add all the numbers together, and all the variables
u!=0/1
u!=0
u∈R
(+2/3)u-16=0
We multiply parentheses
2u^2-16=0
a = 2; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·2·(-16)
Δ = 128
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}$$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{2}}{2*2}=\frac{0-8\sqrt{2}}{4} =-\frac{8\sqrt{2}}{4} =-2\sqrt{2} $$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{2}}{2*2}=\frac{0+8\sqrt{2}}{4} =\frac{8\sqrt{2}}{4} =2\sqrt{2} $
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