27(t)=-5t^2+30t+2

Solution for 27(t)=-5t^2+30t+2 equation:

27(t)=-5t^2+30t+2

We move all terms to the left:

27(t)-(-5t^2+30t+2)=0

We get rid of parentheses

5t^2-30t+27t-2=0

We add all the numbers together, and all the variables

5t^2-3t-2=0

a = 5; b = -3; c = -2;
Δ = b2-4ac
Δ = -32-4·5·(-2)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-7}{2*5}=\frac{-4}{10}
=-2/5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+7}{2*5}=\frac{10}{10}
=1
$