t2+8t+12=0

Solution for t2+8t+12=0 equation:

t2+8t+12=0

We add all the numbers together, and all the variables

t^2+8t+12=0

a = 1; b = 8; c = +12;
Δ = b2-4ac
Δ = 82-4·1·12
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{-(8)-4}{2*1}=\frac{-12}{2}
=-6
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4}{2*1}=\frac{-4}{2}
=-2
$