15=5+27t-16t^2

Solution for 15=5+27t-16t^2 equation:

15=5+27t-16t^2

We move all terms to the left:

15-(5+27t-16t^2)=0

We get rid of parentheses

16t^2-27t-5+15=0

We add all the numbers together, and all the variables

16t^2-27t+10=0

a = 16; b = -27; c = +10;
Δ = b2-4ac
Δ = -272-4·16·10
Δ = 89
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{-(-27)-\sqrt{89}}{2*16}=\frac{27-\sqrt{89}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+\sqrt{89}}{2*16}=\frac{27+\sqrt{89}}{32}
$