X=(2x-7)(3x+1)=(2x-5)(3x+5)

Solution for X=(2x-7)(3x+1)=(2x-5)(3x+5) equation:

X=(2X-7)(3X+1)=(2X-5)(3X+5)

We move all terms to the left:

X-((2X-7)(3X+1))=0

We multiply parentheses ..

-((+6X^2+2X-21X-7))+X=0


We calculate terms in parentheses: -((+6X^2+2X-21X-7)), so:

(+6X^2+2X-21X-7)

We get rid of parentheses

6X^2+2X-21X-7

We add all the numbers together, and all the variables

6X^2-19X-7

Back to the equation:

-(6X^2-19X-7)


We add all the numbers together, and all the variables

X-(6X^2-19X-7)=0

We get rid of parentheses

-6X^2+X+19X+7=0

We add all the numbers together, and all the variables

-6X^2+20X+7=0

a = -6; b = 20; c = +7;
Δ = b2-4ac
Δ = 202-4·(-6)·7
Δ = 568
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{568}=\sqrt{4*142}=\sqrt{4}*\sqrt{142}=2\sqrt{142}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{142}}{2*-6}=\frac{-20-2\sqrt{142}}{-12}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{142}}{2*-6}=\frac{-20+2\sqrt{142}}{-12}
$