16t^2+2t-3=0

Solution for 16t^2+2t-3=0 equation:

16t^2+2t-3=0

a = 16; b = 2; c = -3;
Δ = b2-4ac
Δ = 22-4·16·(-3)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-14}{2*16}=\frac{-16}{32}
=-1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+14}{2*16}=\frac{12}{32}
=3/8
$