2(t+3)t=18+t

Solution for 2(t+3)t=18+t equation:

2(t+3)t=18+t

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

2t^2+6t-(t+18)=0

We get rid of parentheses

2t^2+6t-t-18=0

We add all the numbers together, and all the variables

2t^2+5t-18=0

a = 2; b = 5; c = -18;
Δ = b2-4ac
Δ = 52-4·2·(-18)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-13}{2*2}=\frac{-18}{4}
=-4+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+13}{2*2}=\frac{8}{4}
=2
$