6t^2-48t+42=0

Solution for 6t^2-48t+42=0 equation:

6t^2-48t+42=0

a = 6; b = -48; c = +42;
Δ = b2-4ac
Δ = -482-4·6·42
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-36}{2*6}=\frac{12}{12}
=1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+36}{2*6}=\frac{84}{12}
=7
$