6t^2+24t-10=0

Solution for 6t^2+24t-10=0 equation:

6t^2+24t-10=0

a = 6; b = 24; c = -10;
Δ = b2-4ac
Δ = 242-4·6·(-10)
Δ = 816
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:

$\sqrt{\Delta}=\sqrt{816}=\sqrt{16*51}=\sqrt{16}*\sqrt{51}=4\sqrt{51}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-4\sqrt{51}}{2*6}=\frac{-24-4\sqrt{51}}{12}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+4\sqrt{51}}{2*6}=\frac{-24+4\sqrt{51}}{12}
$