16t^2+30t-13=0

Solution for 16t^2+30t-13=0 equation:

16t^2+30t-13=0

a = 16; b = 30; c = -13;
Δ = b2-4ac
Δ = 302-4·16·(-13)
Δ = 1732
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:

$\sqrt{\Delta}=\sqrt{1732}=\sqrt{4*433}=\sqrt{4}*\sqrt{433}=2\sqrt{433}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-2\sqrt{433}}{2*16}=\frac{-30-2\sqrt{433}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+2\sqrt{433}}{2*16}=\frac{-30+2\sqrt{433}}{32}
$