D=t^2-8t+7

Solution for D=t^2-8t+7 equation:

=D^2-8D+7

We move all terms to the left:

-(D^2-8D+7)=0

We get rid of parentheses

-D^2+8D-7=0

We add all the numbers together, and all the variables

-1D^2+8D-7=0

a = -1; b = 8; c = -7;
Δ = b2-4ac
Δ = 82-4·(-1)·(-7)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-6}{2*-1}=\frac{-14}{-2}
=+7
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+6}{2*-1}=\frac{-2}{-2}
=1
$