-5t^2+6t+3,2=0

Solution for -5t^2+6t+3,2=0 equation:

-5t^2+6t+3.2=0

a = -5; b = 6; c = +3.2;
Δ = b2-4ac
Δ = 62-4·(-5)·3.2
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-10}{2*-5}=\frac{-16}{-10}
=1+3/5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+10}{2*-5}=\frac{4}{-10}
=-2/5
$