27t^2-15t-28=0

Solution for 27t^2-15t-28=0 equation:

27t^2-15t-28=0

a = 27; b = -15; c = -28;
Δ = b2-4ac
Δ = -152-4·27·(-28)
Δ = 3249
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3249}=57$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-57}{2*27}=\frac{-42}{54}
=-7/9
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+57}{2*27}=\frac{72}{54}
=1+1/3
$