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

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

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

We use the square of the difference formula

49t^2-49=0

a = 49; b = 0; c = -49;
Δ = b2-4ac
Δ = 02-4·49·(-49)
Δ = 9604
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9604}=98$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-98}{2*49}=\frac{-98}{98}
=-1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+98}{2*49}=\frac{98}{98}
=1
$