3x^2+3x=17-4x/3

Solution for 3x^2+3x=17-4x/3 equation:

3x^2+3x=17-4x/3

We move all terms to the left:

3x^2+3x-(17-4x/3)=0

We add all the numbers together, and all the variables

3x^2+3x-(-4x/3+17)=0

We get rid of parentheses

3x^2+3x+4x/3-17=0

We multiply all the terms by the denominator


3x^2*3+3x*3+4x-17*3=0

We add all the numbers together, and all the variables

3x^2*3+4x+3x*3-51=0

Wy multiply elements

9x^2+4x+9x-51=0

We add all the numbers together, and all the variables

9x^2+13x-51=0

a = 9; b = 13; c = -51;
Δ = b2-4ac
Δ = 132-4·9·(-51)
Δ = 2005
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-\sqrt{2005}}{2*9}=\frac{-13-\sqrt{2005}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{2005}}{2*9}=\frac{-13+\sqrt{2005}}{18}
$