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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-7x^2+21x+4x-12)-(3(2x-5))=0


We calculate terms in parentheses: -(3(2x-5)), so:

3(2x-5)

We multiply parentheses

6x-15

Back to the equation:

-(6x-15)


We get rid of parentheses

-7x^2+21x+4x-6x-12+15=0

We add all the numbers together, and all the variables

-7x^2+19x+3=0

a = -7; b = 19; c = +3;
Δ = b2-4ac
Δ = 192-4·(-7)·3
Δ = 445
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{-(19)-\sqrt{445}}{2*-7}=\frac{-19-\sqrt{445}}{-14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{445}}{2*-7}=\frac{-19+\sqrt{445}}{-14}
$