33=t^2+4t+22

Solution for 33=t^2+4t+22 equation:

33=t^2+4t+22

We move all terms to the left:

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

We get rid of parentheses

-t^2-4t-22+33=0

We add all the numbers together, and all the variables

-1t^2-4t+11=0

a = -1; b = -4; c = +11;
Δ = b2-4ac
Δ = -42-4·(-1)·11
Δ = 60
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{15}}{2*-1}=\frac{4-2\sqrt{15}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{15}}{2*-1}=\frac{4+2\sqrt{15}}{-2}
$