(1+-1t)(5+t)=0

Solution for (1+-1t)(5+t)=0 equation:

(1+-1t)(5+t)=0

We add all the numbers together, and all the variables

(-1t)(t+5)=0

We multiply parentheses ..

(-1t^2-5t)=0

We get rid of parentheses

-1t^2-5t=0

a = -1; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·(-1)·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*-1}=\frac{0}{-2}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*-1}=\frac{10}{-2}
=-5
$