3(t)=-5*t^2+14*t+3

Solution for 3(t)=-5*t^2+14*t+3 equation:

3(t)=-5t^2+14t+3

We move all terms to the left:

3(t)-(-5t^2+14t+3)=0

We get rid of parentheses

5t^2-14t+3t-3=0

We add all the numbers together, and all the variables

5t^2-11t-3=0

a = 5; b = -11; c = -3;
Δ = b2-4ac
Δ = -112-4·5·(-3)
Δ = 181
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{-(-11)-\sqrt{181}}{2*5}=\frac{11-\sqrt{181}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{181}}{2*5}=\frac{11+\sqrt{181}}{10}
$