3x(2x-5)=13-2(x+2)

Solution for 3x(2x-5)=13-2(x+2) equation:

3x(2x-5)=13-2(x+2)

We move all terms to the left:

3x(2x-5)-(13-2(x+2))=0

We multiply parentheses

6x^2-15x-(13-2(x+2))=0


We calculate terms in parentheses: -(13-2(x+2)), so:

13-2(x+2)

determiningTheFunctionDomain
-2(x+2)+13

We multiply parentheses

-2x-4+13

We add all the numbers together, and all the variables

-2x+9

Back to the equation:

-(-2x+9)


We get rid of parentheses

6x^2-15x+2x-9=0

We add all the numbers together, and all the variables

6x^2-13x-9=0

a = 6; b = -13; c = -9;
Δ = b2-4ac
Δ = -132-4·6·(-9)
Δ = 385
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{385}}{2*6}=\frac{13-\sqrt{385}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{385}}{2*6}=\frac{13+\sqrt{385}}{12}
$