(7t-3)(7t+3)=0

Solution for (7t-3)(7t+3)=0 equation:

(7t-3)(7t+3)=0

We use the square of the difference formula

49t^2-9=0

a = 49; b = 0; c = -9;
Δ = b2-4ac
Δ = 02-4·49·(-9)
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-42}{2*49}=\frac{-42}{98}
=-3/7
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+42}{2*49}=\frac{42}{98}
=3/7
$