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

Solution for 3+14t-5(t^2)=0. equation:

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

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

a = -5; b = 14; c = +3;
Δ = b2-4ac
Δ = 142-4·(-5)·3
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-16}{2*-5}=\frac{-30}{-10}
=+3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+16}{2*-5}=\frac{2}{-10}
=-1/5
$