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

Solution for (x+1)(2x+5)=(4x-7)(2x-5) equation:

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

We move all terms to the left:

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

We multiply parentheses ..

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


We calculate terms in parentheses: -((4x-7)(2x-5)), so:

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

We multiply parentheses ..

(+8x^2-20x-14x+35)

We get rid of parentheses

8x^2-20x-14x+35

We add all the numbers together, and all the variables

8x^2-34x+35

Back to the equation:

-(8x^2-34x+35)


We get rid of parentheses

2x^2-8x^2+5x+2x+34x+5-35=0

We add all the numbers together, and all the variables

-6x^2+41x-30=0

a = -6; b = 41; c = -30;
Δ = b2-4ac
Δ = 412-4·(-6)·(-30)
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{-(41)-31}{2*-6}=\frac{-72}{-12}
=+6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+31}{2*-6}=\frac{-10}{-12}
=5/6
$