t^2+2t+3=18

Solution for t^2+2t+3=18 equation:

t^2+2t+3=18

We move all terms to the left:

t^2+2t+3-(18)=0

We add all the numbers together, and all the variables

t^2+2t-15=0

a = 1; b = 2; c = -15;
Δ = b2-4ac
Δ = 22-4·1·(-15)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-8}{2*1}=\frac{-10}{2}
=-5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+8}{2*1}=\frac{6}{2}
=3
$