0=4t+22t2

Solution for 0=4t+22t2 equation:

0=4t+22t^2

We move all terms to the left:

0-(4t+22t^2)=0

We add all the numbers together, and all the variables

-(4t+22t^2)=0

We get rid of parentheses

-22t^2-4t=0

a = -22; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·(-22)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*-22}=\frac{0}{-44}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*-22}=\frac{8}{-44}
=-2/11
$