(2x-5)(-8x-7)=(3x+7)(8x-20)

Solution for (2x-5)(-8x-7)=(3x+7)(8x-20) equation:

(2x-5)(-8x-7)=(3x+7)(8x-20)

We move all terms to the left:

(2x-5)(-8x-7)-((3x+7)(8x-20))=0

We multiply parentheses ..

(-16x^2-14x+40x+35)-((3x+7)(8x-20))=0


We calculate terms in parentheses: -((3x+7)(8x-20)), so:

(3x+7)(8x-20)

We multiply parentheses ..

(+24x^2-60x+56x-140)

We get rid of parentheses

24x^2-60x+56x-140

We add all the numbers together, and all the variables

24x^2-4x-140

Back to the equation:

-(24x^2-4x-140)


We get rid of parentheses

-16x^2-24x^2-14x+40x+4x+35+140=0

We add all the numbers together, and all the variables

-40x^2+30x+175=0

a = -40; b = 30; c = +175;
Δ = b2-4ac
Δ = 302-4·(-40)·175
Δ = 28900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{28900}=170$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-170}{2*-40}=\frac{-200}{-80}
=2+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+170}{2*-40}=\frac{140}{-80}
=-1+3/4
$