-5t^2+8t-3=0

Solution for -5t^2+8t-3=0 equation:

-5t^2+8t-3=0

a = -5; b = 8; c = -3;
Δ = b2-4ac
Δ = 82-4·(-5)·(-3)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2}{2*-5}=\frac{-10}{-10}
=1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2}{2*-5}=\frac{-6}{-10}
=3/5
$