2/3u^2+5=17.

Solution for 2/3u^2+5=17. equation:

2/3u^2+5=17.

We move all terms to the left:

2/3u^2+5-(17.)=0


Domain of the equation: 3u^2!=0

u^2!=0/3

u^2!=√0

u!=0

u∈R


We add all the numbers together, and all the variables

2/3u^2+5-17=0

We add all the numbers together, and all the variables

2/3u^2-12=0

We multiply all the terms by the denominator


-12*3u^2+2=0

Wy multiply elements

-36u^2+2=0

a = -36; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-36)·2
Δ = 288
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:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{2}}{2*-36}=\frac{0-12\sqrt{2}}{-72}
=-\frac{12\sqrt{2}}{-72}
=-\frac{\sqrt{2}}{-6}
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{2}}{2*-36}=\frac{0+12\sqrt{2}}{-72}
=\frac{12\sqrt{2}}{-72}
=\frac{\sqrt{2}}{-6}
$