t2+23t+22=0

Solution for t2+23t+22=0 equation:

t2+23t+22=0

We add all the numbers together, and all the variables

t^2+23t+22=0

a = 1; b = 23; c = +22;
Δ = b2-4ac
Δ = 232-4·1·22
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-21}{2*1}=\frac{-44}{2}
=-22
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+21}{2*1}=\frac{-2}{2}
=-1
$