45(119)=16t^2+45

Solution for 45(119)=16t^2+45 equation:

45(119)=16t^2+45

We move all terms to the left:

45(119)-(16t^2+45)=0

We get rid of parentheses

-16t^2-45+45119=0

We add all the numbers together, and all the variables

-16t^2+45074=0

a = -16; b = 0; c = +45074;
Δ = b2-4ac
Δ = 02-4·(-16)·45074
Δ = 2884736
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:

$\sqrt{\Delta}=\sqrt{2884736}=\sqrt{64*45074}=\sqrt{64}*\sqrt{45074}=8\sqrt{45074}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{45074}}{2*-16}=\frac{0-8\sqrt{45074}}{-32}
=-\frac{8\sqrt{45074}}{-32}
=-\frac{\sqrt{45074}}{-4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{45074}}{2*-16}=\frac{0+8\sqrt{45074}}{-32}
=\frac{8\sqrt{45074}}{-32}
=\frac{\sqrt{45074}}{-4}
$