3t^2-20t+21=0

Solution for 3t^2-20t+21=0 equation:

3t^2-20t+21=0

a = 3; b = -20; c = +21;
Δ = b2-4ac
Δ = -202-4·3·21
Δ = 148
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:

$\sqrt{\Delta}=\sqrt{148}=\sqrt{4*37}=\sqrt{4}*\sqrt{37}=2\sqrt{37}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{37}}{2*3}=\frac{20-2\sqrt{37}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{37}}{2*3}=\frac{20+2\sqrt{37}}{6}
$