2t^2-24t+21=0

Solution for 2t^2-24t+21=0 equation:

2t^2-24t+21=0

a = 2; b = -24; c = +21;
Δ = b2-4ac
Δ = -242-4·2·21
Δ = 408
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:

$\sqrt{\Delta}=\sqrt{408}=\sqrt{4*102}=\sqrt{4}*\sqrt{102}=2\sqrt{102}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-2\sqrt{102}}{2*2}=\frac{24-2\sqrt{102}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+2\sqrt{102}}{2*2}=\frac{24+2\sqrt{102}}{4}
$