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

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

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

We move all terms to the left:

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

We use the square of the difference formula

x^2-((2x+7)(x-4))-16=0

We multiply parentheses ..

x^2-((+2x^2-8x+7x-28))-16=0


We calculate terms in parentheses: -((+2x^2-8x+7x-28)), so:

(+2x^2-8x+7x-28)

We get rid of parentheses

2x^2-8x+7x-28

We add all the numbers together, and all the variables

2x^2-1x-28

Back to the equation:

-(2x^2-1x-28)


We get rid of parentheses

x^2-2x^2+1x+28-16=0

We add all the numbers together, and all the variables

-1x^2+x+12=0

a = -1; b = 1; c = +12;
Δ = b2-4ac
Δ = 12-4·(-1)·12
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-7}{2*-1}=\frac{-8}{-2}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+7}{2*-1}=\frac{6}{-2}
=-3
$