6t^2-48t+42=16

Solution for 6t^2-48t+42=16 equation:

6t^2-48t+42=16

We move all terms to the left:

6t^2-48t+42-(16)=0

We add all the numbers together, and all the variables

6t^2-48t+26=0

a = 6; b = -48; c = +26;
Δ = b2-4ac
Δ = -482-4·6·26
Δ = 1680
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:

$\sqrt{\Delta}=\sqrt{1680}=\sqrt{16*105}=\sqrt{16}*\sqrt{105}=4\sqrt{105}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-4\sqrt{105}}{2*6}=\frac{48-4\sqrt{105}}{12}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+4\sqrt{105}}{2*6}=\frac{48+4\sqrt{105}}{12}
$