(2x-1)(3x+4)=7(2-x)

Solution for (2x-1)(3x+4)=7(2-x) equation:

(2x-1)(3x+4)=7(2-x)

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+6x^2+8x-3x-4)-(7(-1x+2))=0


We calculate terms in parentheses: -(7(-1x+2)), so:

7(-1x+2)

We multiply parentheses

-7x+14

Back to the equation:

-(-7x+14)


We get rid of parentheses

6x^2+8x-3x+7x-4-14=0

We add all the numbers together, and all the variables

6x^2+12x-18=0

a = 6; b = 12; c = -18;
Δ = b2-4ac
Δ = 122-4·6·(-18)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-24}{2*6}=\frac{-36}{12}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+24}{2*6}=\frac{12}{12}
=1
$