0=-16t^2-23t+130

Solution for 0=-16t^2-23t+130 equation:

0=-16t^2-23t+130

We move all terms to the left:

0-(-16t^2-23t+130)=0

We add all the numbers together, and all the variables

-(-16t^2-23t+130)=0

We get rid of parentheses

16t^2+23t-130=0

a = 16; b = 23; c = -130;
Δ = b2-4ac
Δ = 232-4·16·(-130)
Δ = 8849
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-\sqrt{8849}}{2*16}=\frac{-23-\sqrt{8849}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+\sqrt{8849}}{2*16}=\frac{-23+\sqrt{8849}}{32}
$