t^2-48t+480=0

Solution for t^2-48t+480=0 equation:

t^2-48t+480=0

a = 1; b = -48; c = +480;
Δ = b2-4ac
Δ = -482-4·1·480
Δ = 384
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:

$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-8\sqrt{6}}{2*1}=\frac{48-8\sqrt{6}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+8\sqrt{6}}{2*1}=\frac{48+8\sqrt{6}}{2}
$