3t^2-22t+21=0

Solution for 3t^2-22t+21=0 equation:

3t^2-22t+21=0

a = 3; b = -22; c = +21;
Δ = b2-4ac
Δ = -222-4·3·21
Δ = 232
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:

$\sqrt{\Delta}=\sqrt{232}=\sqrt{4*58}=\sqrt{4}*\sqrt{58}=2\sqrt{58}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{58}}{2*3}=\frac{22-2\sqrt{58}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{58}}{2*3}=\frac{22+2\sqrt{58}}{6}
$