0=16t^2-27t+10

Solution for 0=16t^2-27t+10 equation:

0=16t^2-27t+10

We move all terms to the left:

0-(16t^2-27t+10)=0

We add all the numbers together, and all the variables

-(16t^2-27t+10)=0

We get rid of parentheses

-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}
$