(7t+5)(7t-5)=0

Solution for (7t+5)(7t-5)=0 equation:

(7t+5)(7t-5)=0

We use the square of the difference formula

49t^2-25=0

a = 49; b = 0; c = -25;
Δ = b2-4ac
Δ = 02-4·49·(-25)
Δ = 4900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4900}=70$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-70}{2*49}=\frac{-70}{98}
=-5/7
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+70}{2*49}=\frac{70}{98}
=5/7
$