27(.06t)=-16t^2+96t+112

Solution for 27(.06t)=-16t^2+96t+112 equation:

27(.06t)=-16t^2+96t+112

We move all terms to the left:

27(.06t)-(-16t^2+96t+112)=0

We add all the numbers together, and all the variables

-(-16t^2+96t+112)+27(+.06t)=0

We multiply parentheses

-(-16t^2+96t+112)+27t=0

We get rid of parentheses

16t^2-96t+27t-112=0

We add all the numbers together, and all the variables

16t^2-69t-112=0

a = 16; b = -69; c = -112;
Δ = b2-4ac
Δ = -692-4·16·(-112)
Δ = 11929
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{-(-69)-\sqrt{11929}}{2*16}=\frac{69-\sqrt{11929}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-69)+\sqrt{11929}}{2*16}=\frac{69+\sqrt{11929}}{32}
$