(2t+3)(2t-3)=0

Solution for (2t+3)(2t-3)=0 equation:

(2t+3)(2t-3)=0

We use the square of the difference formula

4t^2-9=0

a = 4; b = 0; c = -9;
Δ = b2-4ac
Δ = 02-4·4·(-9)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12}{2*4}=\frac{-12}{8}
=-1+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*4}=\frac{12}{8}
=1+1/2
$