4=-16t^2+45+5

Solution for 4=-16t^2+45+5 equation:

4=-16t^2+45+5

We move all terms to the left:

4-(-16t^2+45+5)=0

We get rid of parentheses

16t^2-45-5+4=0

We add all the numbers together, and all the variables

16t^2-46=0

a = 16; b = 0; c = -46;
Δ = b2-4ac
Δ = 02-4·16·(-46)
Δ = 2944
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:

$\sqrt{\Delta}=\sqrt{2944}=\sqrt{64*46}=\sqrt{64}*\sqrt{46}=8\sqrt{46}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{46}}{2*16}=\frac{0-8\sqrt{46}}{32}
=-\frac{8\sqrt{46}}{32}
=-\frac{\sqrt{46}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{46}}{2*16}=\frac{0+8\sqrt{46}}{32}
=\frac{8\sqrt{46}}{32}
=\frac{\sqrt{46}}{4}
$