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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

-2(3x-4)-(9+5x(-1x+3))+7=0

We multiply parentheses

-6x-(9+5x(-1x+3))+8+7=0


We calculate terms in parentheses: -(9+5x(-1x+3)), so:

9+5x(-1x+3)

determiningTheFunctionDomain
5x(-1x+3)+9

We multiply parentheses

-5x^2+15x+9

Back to the equation:

-(-5x^2+15x+9)


We add all the numbers together, and all the variables

-(-5x^2+15x+9)-6x+15=0

We get rid of parentheses

5x^2-15x-6x-9+15=0

We add all the numbers together, and all the variables

5x^2-21x+6=0

a = 5; b = -21; c = +6;
Δ = b2-4ac
Δ = -212-4·5·6
Δ = 321
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{-(-21)-\sqrt{321}}{2*5}=\frac{21-\sqrt{321}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+\sqrt{321}}{2*5}=\frac{21+\sqrt{321}}{10}
$