(4t+6)(4t-6)=0

Solution for (4t+6)(4t-6)=0 equation:

(4t+6)(4t-6)=0

We use the square of the difference formula

16t^2-36=0

a = 16; b = 0; c = -36;
Δ = b2-4ac
Δ = 02-4·16·(-36)
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48}{2*16}=\frac{-48}{32}
=-1+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48}{2*16}=\frac{48}{32}
=1+1/2
$