t^2+8t+2=22

Solution for t^2+8t+2=22 equation:

t^2+8t+2=22

We move all terms to the left:

t^2+8t+2-(22)=0

We add all the numbers together, and all the variables

t^2+8t-20=0

a = 1; b = 8; c = -20;
Δ = b2-4ac
Δ = 82-4·1·(-20)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-12}{2*1}=\frac{-20}{2}
=-10
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+12}{2*1}=\frac{4}{2}
=2
$