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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We multiply parentheses ..

-((+20x^2-35x+12x-21))+4x-7=0


We calculate terms in parentheses: -((+20x^2-35x+12x-21)), so:

(+20x^2-35x+12x-21)

We get rid of parentheses

20x^2-35x+12x-21

We add all the numbers together, and all the variables

20x^2-23x-21

Back to the equation:

-(20x^2-23x-21)


We add all the numbers together, and all the variables

4x-(20x^2-23x-21)-7=0

We get rid of parentheses

-20x^2+4x+23x+21-7=0

We add all the numbers together, and all the variables

-20x^2+27x+14=0

a = -20; b = 27; c = +14;
Δ = b2-4ac
Δ = 272-4·(-20)·14
Δ = 1849
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1849}=43$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-43}{2*-20}=\frac{-70}{-40}
=1+3/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+43}{2*-20}=\frac{16}{-40}
=-2/5
$