-16t2+17t+-4=0

Solution for -16t2+17t+-4=0 equation:

-16t^2+17t+-4=0

We add all the numbers together, and all the variables

-16t^2+17t=0

a = -16; b = 17; c = 0;
Δ = b2-4ac
Δ = 172-4·(-16)·0
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-17}{2*-16}=\frac{-34}{-32}
=1+1/16
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+17}{2*-16}=\frac{0}{-32}
=0
$