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

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

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

We move all terms to the left:

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

We use the square of the difference formula

4x^2-((3x+7))-25=0


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

(3x+7)

We get rid of parentheses

3x+7

Back to the equation:

-(3x+7)


We get rid of parentheses

4x^2-3x-7-25=0

We add all the numbers together, and all the variables

4x^2-3x-32=0

a = 4; b = -3; c = -32;
Δ = b2-4ac
Δ = -32-4·4·(-32)
Δ = 521
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{-(-3)-\sqrt{521}}{2*4}=\frac{3-\sqrt{521}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{521}}{2*4}=\frac{3+\sqrt{521}}{8}
$