11t(2t+4)=0

Solution for 11t(2t+4)=0 equation:

11t(2t+4)=0

We multiply parentheses

22t^2+44t=0

a = 22; b = 44; c = 0;
Δ = b2-4ac
Δ = 442-4·22·0
Δ = 1936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1936}=44$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(44)-44}{2*22}=\frac{-88}{44}
=-2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(44)+44}{2*22}=\frac{0}{44}
=0
$