20=6t2+63t+4

Solution for 20=6t2+63t+4 equation:

20=6t^2+63t+4

We move all terms to the left:

20-(6t^2+63t+4)=0

We get rid of parentheses

-6t^2-63t-4+20=0

We add all the numbers together, and all the variables

-6t^2-63t+16=0

a = -6; b = -63; c = +16;
Δ = b2-4ac
Δ = -632-4·(-6)·16
Δ = 4353
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{-(-63)-\sqrt{4353}}{2*-6}=\frac{63-\sqrt{4353}}{-12}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-63)+\sqrt{4353}}{2*-6}=\frac{63+\sqrt{4353}}{-12}
$