16t^2+32t-9=0

Solution for 16t^2+32t-9=0 equation:

16t^2+32t-9=0

a = 16; b = 32; c = -9;
Δ = b2-4ac
Δ = 322-4·16·(-9)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-40}{2*16}=\frac{-72}{32}
=-2+1/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+40}{2*16}=\frac{8}{32}
=1/4
$