8t^2+18t+9=8t+12t+6t+9

Solution for 8t^2+18t+9=8t+12t+6t+9 equation:

8t^2+18t+9=8t+12t+6t+9

We move all terms to the left:

8t^2+18t+9-(8t+12t+6t+9)=0

We add all the numbers together, and all the variables

8t^2+18t-(26t+9)+9=0

We get rid of parentheses

8t^2+18t-26t-9+9=0

We add all the numbers together, and all the variables

8t^2-8t=0

a = 8; b = -8; c = 0;
Δ = b2-4ac
Δ = -82-4·8·0
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{-(-8)-8}{2*8}=\frac{0}{16}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8}{2*8}=\frac{16}{16}
=1
$