-15t^2-23t+28=0

Solution for -15t^2-23t+28=0 equation:

-15t^2-23t+28=0

a = -15; b = -23; c = +28;
Δ = b2-4ac
Δ = -232-4·(-15)·28
Δ = 2209
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2209}=47$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-47}{2*-15}=\frac{-24}{-30}
=4/5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+47}{2*-15}=\frac{70}{-30}
=-2+1/3
$