1=(t^2)-8t+16

Solution for 1=(t^2)-8t+16 equation:

1=(t^2)-8t+16

We move all terms to the left:

1-((t^2)-8t+16)=0

We get rid of parentheses

-t^2+8t-16+1=0

We add all the numbers together, and all the variables

-1t^2+8t-15=0

a = -1; b = 8; c = -15;
Δ = b2-4ac
Δ = 82-4·(-1)·(-15)
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{-(8)-2}{2*-1}=\frac{-10}{-2}
=+5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2}{2*-1}=\frac{-6}{-2}
=+3
$