(6x-5)(x+1)=(2x-7)(2x-1)

Solution for (6x-5)(x+1)=(2x-7)(2x-1) equation:

(6x-5)(x+1)=(2x-7)(2x-1)

We move all terms to the left:

(6x-5)(x+1)-((2x-7)(2x-1))=0

We multiply parentheses ..

(+6x^2+6x-5x-5)-((2x-7)(2x-1))=0


We calculate terms in parentheses: -((2x-7)(2x-1)), so:

(2x-7)(2x-1)

We multiply parentheses ..

(+4x^2-2x-14x+7)

We get rid of parentheses

4x^2-2x-14x+7

We add all the numbers together, and all the variables

4x^2-16x+7

Back to the equation:

-(4x^2-16x+7)


We get rid of parentheses

6x^2-4x^2+6x-5x+16x-5-7=0

We add all the numbers together, and all the variables

2x^2+17x-12=0

a = 2; b = 17; c = -12;
Δ = b2-4ac
Δ = 172-4·2·(-12)
Δ = 385
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-\sqrt{385}}{2*2}=\frac{-17-\sqrt{385}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+\sqrt{385}}{2*2}=\frac{-17+\sqrt{385}}{4}
$