t(-3t-28)=0

Solution for t(-3t-28)=0 equation:

t(-3t-28)=0

We multiply parentheses

-3t^2-28t=0

a = -3; b = -28; c = 0;
Δ = b2-4ac
Δ = -282-4·(-3)·0
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-28}{2*-3}=\frac{0}{-6}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+28}{2*-3}=\frac{56}{-6}
=-9+1/3
$