45(129)=16t^2+45

Solution for 45(129)=16t^2+45 equation:

45(129)=16t^2+45

We move all terms to the left:

45(129)-(16t^2+45)=0

We get rid of parentheses

-16t^2-45+45129=0

We add all the numbers together, and all the variables

-16t^2+45084=0

a = -16; b = 0; c = +45084;
Δ = b2-4ac
Δ = 02-4·(-16)·45084
Δ = 2885376
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:

$\sqrt{\Delta}=\sqrt{2885376}=\sqrt{73984*39}=\sqrt{73984}*\sqrt{39}=272\sqrt{39}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-272\sqrt{39}}{2*-16}=\frac{0-272\sqrt{39}}{-32}
=-\frac{272\sqrt{39}}{-32}
=-\frac{17\sqrt{39}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+272\sqrt{39}}{2*-16}=\frac{0+272\sqrt{39}}{-32}
=\frac{272\sqrt{39}}{-32}
=\frac{17\sqrt{39}}{-2}
$