-15t^2-4t+3=0

Solution for -15t^2-4t+3=0 equation:

-15t^2-4t+3=0

a = -15; b = -4; c = +3;
Δ = b2-4ac
Δ = -42-4·(-15)·3
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-14}{2*-15}=\frac{-10}{-30}
=1/3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+14}{2*-15}=\frac{18}{-30}
=-3/5
$