t^2-24t-10=0

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

t^2-24t-10=0

a = 1; b = -24; c = -10;
Δ = b2-4ac
Δ = -242-4·1·(-10)
Δ = 616
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:

$\sqrt{\Delta}=\sqrt{616}=\sqrt{4*154}=\sqrt{4}*\sqrt{154}=2\sqrt{154}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-2\sqrt{154}}{2*1}=\frac{24-2\sqrt{154}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+2\sqrt{154}}{2*1}=\frac{24+2\sqrt{154}}{2}
$