2/3x^2+5=18

Solution for 2/3x^2+5=18 equation:

2/3x^2+5=18

We move all terms to the left:

2/3x^2+5-(18)=0


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

x^2!=0/3

x^2!=√0

x!=0

x∈R


We add all the numbers together, and all the variables

2/3x^2-13=0

We multiply all the terms by the denominator


-13*3x^2+2=0

Wy multiply elements

-39x^2+2=0

a = -39; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-39)·2
Δ = 312
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{312}=\sqrt{4*78}=\sqrt{4}*\sqrt{78}=2\sqrt{78}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{78}}{2*-39}=\frac{0-2\sqrt{78}}{-78}
=-\frac{2\sqrt{78}}{-78}
=-\frac{\sqrt{78}}{-39}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{78}}{2*-39}=\frac{0+2\sqrt{78}}{-78}
=\frac{2\sqrt{78}}{-78}
=\frac{\sqrt{78}}{-39}
$