28(t)=-9.2t^2+18t+32

Solution for 28(t)=-9.2t^2+18t+32 equation:

28(t)=-9.2t^2+18t+32

We move all terms to the left:

28(t)-(-9.2t^2+18t+32)=0

We get rid of parentheses

9.2t^2-18t+28t-32=0

We add all the numbers together, and all the variables

9.2t^2+10t-32=0

a = 9.2; b = 10; c = -32;
Δ = b2-4ac
Δ = 102-4·9.2·(-32)
Δ = 1277.6
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{-(10)-\sqrt{1277.6}}{2*9.2}=\frac{-10-\sqrt{1277.6}}{18.4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+\sqrt{1277.6}}{2*9.2}=\frac{-10+\sqrt{1277.6}}{18.4}
$