36=-4t^2-t+45

Solution for 36=-4t^2-t+45 equation:

36=-4t^2-t+45

We move all terms to the left:

36-(-4t^2-t+45)=0

We get rid of parentheses

4t^2+t-45+36=0

We add all the numbers together, and all the variables

4t^2+t-9=0

a = 4; b = 1; c = -9;
Δ = b2-4ac
Δ = 12-4·4·(-9)
Δ = 145
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{145}}{2*4}=\frac{-1-\sqrt{145}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{145}}{2*4}=\frac{-1+\sqrt{145}}{8}
$