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