6t(2t+1)-9=t-(t+2)

Solution for 6t(2t+1)-9=t-(t+2) equation:

6t(2t+1)-9=t-(t+2)

We move all terms to the left:

6t(2t+1)-9-(t-(t+2))=0

We multiply parentheses

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


We calculate terms in parentheses: -(t-(t+2)), so:

t-(t+2)

We get rid of parentheses

t-t-2

We add all the numbers together, and all the variables

-2

Back to the equation:

-(-2)


We add all the numbers together, and all the variables

12t^2+6t-7=0

a = 12; b = 6; c = -7;
Δ = b2-4ac
Δ = 62-4·12·(-7)
Δ = 372
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{372}=\sqrt{4*93}=\sqrt{4}*\sqrt{93}=2\sqrt{93}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{93}}{2*12}=\frac{-6-2\sqrt{93}}{24}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{93}}{2*12}=\frac{-6+2\sqrt{93}}{24}
$