39=(2x+3)(2x+5)

Solution for 39=(2x+3)(2x+5) equation:

39=(2x+3)(2x+5)

We move all terms to the left:

39-((2x+3)(2x+5))=0

We multiply parentheses ..

-((+4x^2+10x+6x+15))+39=0


We calculate terms in parentheses: -((+4x^2+10x+6x+15)), so:

(+4x^2+10x+6x+15)

We get rid of parentheses

4x^2+10x+6x+15

We add all the numbers together, and all the variables

4x^2+16x+15

Back to the equation:

-(4x^2+16x+15)


We get rid of parentheses

-4x^2-16x-15+39=0

We add all the numbers together, and all the variables

-4x^2-16x+24=0

a = -4; b = -16; c = +24;
Δ = b2-4ac
Δ = -162-4·(-4)·24
Δ = 640
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:

$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-8\sqrt{10}}{2*-4}=\frac{16-8\sqrt{10}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+8\sqrt{10}}{2*-4}=\frac{16+8\sqrt{10}}{-8}
$