(2x+3)(4x+5)=(7x-6)(2x+3)

Solution for (2x+3)(4x+5)=(7x-6)(2x+3) equation:

(2x+3)(4x+5)=(7x-6)(2x+3)

We move all terms to the left:

(2x+3)(4x+5)-((7x-6)(2x+3))=0

We multiply parentheses ..

(+8x^2+10x+12x+15)-((7x-6)(2x+3))=0


We calculate terms in parentheses: -((7x-6)(2x+3)), so:

(7x-6)(2x+3)

We multiply parentheses ..

(+14x^2+21x-12x-18)

We get rid of parentheses

14x^2+21x-12x-18

We add all the numbers together, and all the variables

14x^2+9x-18

Back to the equation:

-(14x^2+9x-18)


We get rid of parentheses

8x^2-14x^2+10x+12x-9x+15+18=0

We add all the numbers together, and all the variables

-6x^2+13x+33=0

a = -6; b = 13; c = +33;
Δ = b2-4ac
Δ = 132-4·(-6)·33
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-31}{2*-6}=\frac{-44}{-12}
=3+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+31}{2*-6}=\frac{18}{-12}
=-1+1/2
$