8t-t2=0

Solution for 8t-t2=0 equation:

8t-t2=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{-16}{-2}
=+8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*-1}=\frac{0}{-2}
=0
$