-45=-16t^2+34

Solution for -45=-16t^2+34 equation:

-45=-16t^2+34

We move all terms to the left:

-45-(-16t^2+34)=0

We get rid of parentheses

16t^2-34-45=0

We add all the numbers together, and all the variables

16t^2-79=0

a = 16; b = 0; c = -79;
Δ = b2-4ac
Δ = 02-4·16·(-79)
Δ = 5056
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:

$\sqrt{\Delta}=\sqrt{5056}=\sqrt{64*79}=\sqrt{64}*\sqrt{79}=8\sqrt{79}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{79}}{2*16}=\frac{0-8\sqrt{79}}{32}
=-\frac{8\sqrt{79}}{32}
=-\frac{\sqrt{79}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{79}}{2*16}=\frac{0+8\sqrt{79}}{32}
=\frac{8\sqrt{79}}{32}
=\frac{\sqrt{79}}{4}
$