3x(2x-3)-7=9-3(4x-5)

Solution for 3x(2x-3)-7=9-3(4x-5) equation:

3x(2x-3)-7=9-3(4x-5)

We move all terms to the left:

3x(2x-3)-7-(9-3(4x-5))=0

We multiply parentheses

6x^2-9x-(9-3(4x-5))-7=0


We calculate terms in parentheses: -(9-3(4x-5)), so:

9-3(4x-5)

determiningTheFunctionDomain
-3(4x-5)+9

We multiply parentheses

-12x+15+9

We add all the numbers together, and all the variables

-12x+24

Back to the equation:

-(-12x+24)


We get rid of parentheses

6x^2-9x+12x-24-7=0

We add all the numbers together, and all the variables

6x^2+3x-31=0

a = 6; b = 3; c = -31;
Δ = b2-4ac
Δ = 32-4·6·(-31)
Δ = 753
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{-(3)-\sqrt{753}}{2*6}=\frac{-3-\sqrt{753}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{753}}{2*6}=\frac{-3+\sqrt{753}}{12}
$