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

Solution for -t^2+-8t+15=0 equation:

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

We add all the numbers together, and all the variables

-1t^2-8t=0

a = -1; b = -8; c = 0;
Δ = b2-4ac
Δ = -82-4·(-1)·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8}{2*-1}=\frac{0}{-2}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8}{2*-1}=\frac{16}{-2}
=-8
$