3t^2-25t+8=0

Solution for 3t^2-25t+8=0 equation:

3t^2-25t+8=0

a = 3; b = -25; c = +8;
Δ = b2-4ac
Δ = -252-4·3·8
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-23}{2*3}=\frac{2}{6}
=1/3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+23}{2*3}=\frac{48}{6}
=8
$