(3x^2+2x)/9=5

Solution for (3x^2+2x)/9=5 equation:

(3x^2+2x)/9=5

We move all terms to the left:

(3x^2+2x)/9-(5)=0

We multiply all the terms by the denominator


(3x^2+2x)-5*9=0

We add all the numbers together, and all the variables

(3x^2+2x)-45=0

We get rid of parentheses

3x^2+2x-45=0

a = 3; b = 2; c = -45;
Δ = b2-4ac
Δ = 22-4·3·(-45)
Δ = 544
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:

$\sqrt{\Delta}=\sqrt{544}=\sqrt{16*34}=\sqrt{16}*\sqrt{34}=4\sqrt{34}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4\sqrt{34}}{2*3}=\frac{-2-4\sqrt{34}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4\sqrt{34}}{2*3}=\frac{-2+4\sqrt{34}}{6}
$