2t^2+4t=9t+18

Solution for 2t^2+4t=9t+18 equation:

2t^2+4t=9t+18

We move all terms to the left:

2t^2+4t-(9t+18)=0

We get rid of parentheses

2t^2+4t-9t-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{-8}{4}
=-2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+13}{2*2}=\frac{18}{4}
=4+1/2
$