(t-7)(t-9)=0

Solution for (t-7)(t-9)=0 equation:

(t-7)(t-9)=0

We multiply parentheses ..

(+t^2-9t-7t+63)=0

We get rid of parentheses

t^2-9t-7t+63=0

We add all the numbers together, and all the variables

t^2-16t+63=0

a = 1; b = -16; c = +63;
Δ = b2-4ac
Δ = -162-4·1·63
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2}{2*1}=\frac{14}{2}
=7
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2}{2*1}=\frac{18}{2}
=9
$