0.8t^2-7t=0

Solution for 0.8t^2-7t=0 equation:

0.8t^2-7t=0

a = 0.8; b = -7; c = 0;
Δ = b2-4ac
Δ = -72-4·0.8·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-7}{2*0.8}=\frac{0}{1.6}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+7}{2*0.8}=\frac{14}{1.6}
=8+1/1.3333333333333
$