3t^2-40t+81=0

Solution for 3t^2-40t+81=0 equation:

3t^2-40t+81=0

a = 3; b = -40; c = +81;
Δ = b2-4ac
Δ = -402-4·3·81
Δ = 628
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{628}=\sqrt{4*157}=\sqrt{4}*\sqrt{157}=2\sqrt{157}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-2\sqrt{157}}{2*3}=\frac{40-2\sqrt{157}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+2\sqrt{157}}{2*3}=\frac{40+2\sqrt{157}}{6}
$