t^2+12t+36=04

Solution for t^2+12t+36=04 equation:

t^2+12t+36=04

We move all terms to the left:

t^2+12t+36-(04)=0

We add all the numbers together, and all the variables

t^2+12t+32=0

a = 1; b = 12; c = +32;
Δ = b2-4ac
Δ = 122-4·1·32
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{-(12)-4}{2*1}=\frac{-16}{2}
=-8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4}{2*1}=\frac{-8}{2}
=-4
$