3t^2-42t+48=0

Solution for 3t^2-42t+48=0 equation:

3t^2-42t+48=0

a = 3; b = -42; c = +48;
Δ = b2-4ac
Δ = -422-4·3·48
Δ = 1188
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:

$\sqrt{\Delta}=\sqrt{1188}=\sqrt{36*33}=\sqrt{36}*\sqrt{33}=6\sqrt{33}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-6\sqrt{33}}{2*3}=\frac{42-6\sqrt{33}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+6\sqrt{33}}{2*3}=\frac{42+6\sqrt{33}}{6}
$