6t^2+33t+5=21

Solution for 6t^2+33t+5=21 equation:

6t^2+33t+5=21

We move all terms to the left:

6t^2+33t+5-(21)=0

We add all the numbers together, and all the variables

6t^2+33t-16=0

a = 6; b = 33; c = -16;
Δ = b2-4ac
Δ = 332-4·6·(-16)
Δ = 1473
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-\sqrt{1473}}{2*6}=\frac{-33-\sqrt{1473}}{12}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+\sqrt{1473}}{2*6}=\frac{-33+\sqrt{1473}}{12}
$