(3a-4)(3a+4)=7

Solution for (3a-4)(3a+4)=7 equation:

(3a-4)(3a+4)=7

We move all terms to the left:

(3a-4)(3a+4)-(7)=0

We use the square of the difference formula

9a^2-16-7=0

We add all the numbers together, and all the variables

9a^2-23=0

a = 9; b = 0; c = -23;
Δ = b2-4ac
Δ = 02-4·9·(-23)
Δ = 828
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{828}=\sqrt{36*23}=\sqrt{36}*\sqrt{23}=6\sqrt{23}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{23}}{2*9}=\frac{0-6\sqrt{23}}{18}
=-\frac{6\sqrt{23}}{18}
=-\frac{\sqrt{23}}{3}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{23}}{2*9}=\frac{0+6\sqrt{23}}{18}
=\frac{6\sqrt{23}}{18}
=\frac{\sqrt{23}}{3}
$