-3=-16t^2+30t+4

Solution for -3=-16t^2+30t+4 equation:

-3=-16t^2+30t+4

We move all terms to the left:

-3-(-16t^2+30t+4)=0

We get rid of parentheses

16t^2-30t-4-3=0

We add all the numbers together, and all the variables

16t^2-30t-7=0

a = 16; b = -30; c = -7;
Δ = b2-4ac
Δ = -302-4·16·(-7)
Δ = 1348
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:

$\sqrt{\Delta}=\sqrt{1348}=\sqrt{4*337}=\sqrt{4}*\sqrt{337}=2\sqrt{337}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-2\sqrt{337}}{2*16}=\frac{30-2\sqrt{337}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+2\sqrt{337}}{2*16}=\frac{30+2\sqrt{337}}{32}
$