(2x+3)(2x+4)-9=7

Solution for (2x+3)(2x+4)-9=7 equation:

(2x+3)(2x+4)-9=7

We move all terms to the left:

(2x+3)(2x+4)-9-(7)=0

We add all the numbers together, and all the variables

(2x+3)(2x+4)-16=0

We multiply parentheses ..

(+4x^2+8x+6x+12)-16=0

We get rid of parentheses

4x^2+8x+6x+12-16=0

We add all the numbers together, and all the variables

4x^2+14x-4=0

a = 4; b = 14; c = -4;
Δ = b2-4ac
Δ = 142-4·4·(-4)
Δ = 260
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:

$\sqrt{\Delta}=\sqrt{260}=\sqrt{4*65}=\sqrt{4}*\sqrt{65}=2\sqrt{65}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{65}}{2*4}=\frac{-14-2\sqrt{65}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{65}}{2*4}=\frac{-14+2\sqrt{65}}{8}
$