4=21t^2+8t

Solution for 4=21t^2+8t equation:

4=21t^2+8t

We move all terms to the left:

4-(21t^2+8t)=0

We get rid of parentheses

-21t^2-8t+4=0

a = -21; b = -8; c = +4;
Δ = b2-4ac
Δ = -82-4·(-21)·4
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-20}{2*-21}=\frac{-12}{-42}
=2/7
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+20}{2*-21}=\frac{28}{-42}
=-2/3
$