2x^2+(1/3)x-(10/3)=0

Solution for 2x^2+(1/3)x-(10/3)=0 equation:

2x^2+(1/3)x-(10/3)=0


Domain of the equation: 3)x!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

2x^2+(+1/3)x-(+10/3)=0

We multiply parentheses

2x^2+x^2-(+10/3)=0

We get rid of parentheses

2x^2+x^2-10/3=0

We multiply all the terms by the denominator


2x^2*3+x^2*3-10=0

Wy multiply elements

6x^2+3x^2-10=0

We add all the numbers together, and all the variables

9x^2-10=0

a = 9; b = 0; c = -10;
Δ = b2-4ac
Δ = 02-4·9·(-10)
Δ = 360
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:

$\sqrt{\Delta}=\sqrt{360}=\sqrt{36*10}=\sqrt{36}*\sqrt{10}=6\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{10}}{2*9}=\frac{0-6\sqrt{10}}{18}
=-\frac{6\sqrt{10}}{18}
=-\frac{\sqrt{10}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{10}}{2*9}=\frac{0+6\sqrt{10}}{18}
=\frac{6\sqrt{10}}{18}
=\frac{\sqrt{10}}{3}
$