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

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

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

We move all terms to the left:

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


We calculate terms in parentheses: -(7-(2x^2-3x-4)), so:

7-(2x^2-3x-4)

determiningTheFunctionDomain
-(2x^2-3x-4)+7

We get rid of parentheses

-2x^2+3x+4+7

We add all the numbers together, and all the variables

-2x^2+3x+11

Back to the equation:

-(-2x^2+3x+11)


We get rid of parentheses

5x^2+2x^2-3x-4x-11=0

We add all the numbers together, and all the variables

7x^2-7x-11=0

a = 7; b = -7; c = -11;
Δ = b2-4ac
Δ = -72-4·7·(-11)
Δ = 357
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{-(-7)-\sqrt{357}}{2*7}=\frac{7-\sqrt{357}}{14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{357}}{2*7}=\frac{7+\sqrt{357}}{14}
$