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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+x^2-5x+7x-35)-((2x+1)(x-5))=0


We calculate terms in parentheses: -((2x+1)(x-5)), so:

(2x+1)(x-5)

We multiply parentheses ..

(+2x^2-10x+x-5)

We get rid of parentheses

2x^2-10x+x-5

We add all the numbers together, and all the variables

2x^2-9x-5

Back to the equation:

-(2x^2-9x-5)


We get rid of parentheses

x^2-2x^2-5x+7x+9x-35+5=0

We add all the numbers together, and all the variables

-1x^2+11x-30=0

a = -1; b = 11; c = -30;
Δ = b2-4ac
Δ = 112-4·(-1)·(-30)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-1}{2*-1}=\frac{-12}{-2}
=+6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+1}{2*-1}=\frac{-10}{-2}
=+5
$