-16t^2+158t-40=0

Solution for -16t^2+158t-40=0 equation:

-16t^2+158t-40=0

a = -16; b = 158; c = -40;
Δ = b2-4ac
Δ = 1582-4·(-16)·(-40)
Δ = 22404
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:

$\sqrt{\Delta}=\sqrt{22404}=\sqrt{4*5601}=\sqrt{4}*\sqrt{5601}=2\sqrt{5601}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(158)-2\sqrt{5601}}{2*-16}=\frac{-158-2\sqrt{5601}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(158)+2\sqrt{5601}}{2*-16}=\frac{-158+2\sqrt{5601}}{-32}
$