16t^2-30t+4=0

Solution for 16t^2-30t+4=0 equation:

16t^2-30t+4=0

a = 16; b = -30; c = +4;
Δ = b2-4ac
Δ = -302-4·16·4
Δ = 644
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:

$\sqrt{\Delta}=\sqrt{644}=\sqrt{4*161}=\sqrt{4}*\sqrt{161}=2\sqrt{161}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-2\sqrt{161}}{2*16}=\frac{30-2\sqrt{161}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+2\sqrt{161}}{2*16}=\frac{30+2\sqrt{161}}{32}
$