3^2-2*3+u=(3^3/3*8)(2)+u

Solution for 3^2-2*3+u=(3^3/3*8)(2)+u equation:

3^2-2*3+u=(3^3/3*8)(2)+u

We move all terms to the left:

3^2-2*3+u-((3^3/3*8)(2)+u)=0


Domain of the equation: 3*8)2+u)!=0

u!=0/1

u!=0

u∈R


We add all the numbers together, and all the variables

u-((3^3/3*8)2+u)+3=0

We multiply all the terms by the denominator


u*3*8)2+u)-((3^3+3*3*8)2+u)=0

Wy multiply elements

24u^2*8+72u*8=0

Wy multiply elements

192u^2+576u=0

a = 192; b = 576; c = 0;
Δ = b2-4ac
Δ = 5762-4·192·0
Δ = 331776
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{331776}=576$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(576)-576}{2*192}=\frac{-1152}{384}
=-3
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(576)+576}{2*192}=\frac{0}{384}
=0
$