(2x^2/9)+x=2

Solution for (2x^2/9)+x=2 equation:

(2x^2/9)+x=2

We move all terms to the left:

(2x^2/9)+x-(2)=0

We add all the numbers together, and all the variables

x+(2x^2/9)-2=0

We get rid of parentheses

2x^2/9+x-2=0

We multiply all the terms by the denominator


2x^2+x*9-2*9=0

We add all the numbers together, and all the variables

2x^2+x*9-18=0

Wy multiply elements

2x^2+9x-18=0

a = 2; b = 9; c = -18;
Δ = b2-4ac
Δ = 92-4·2·(-18)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-15}{2*2}=\frac{-24}{4}
=-6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+15}{2*2}=\frac{6}{4}
=1+1/2
$