3t(3t-4)=2(t+8)

Solution for 3t(3t-4)=2(t+8) equation:

3t(3t-4)=2(t+8)

We move all terms to the left:

3t(3t-4)-(2(t+8))=0

We multiply parentheses

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


We calculate terms in parentheses: -(2(t+8)), so:

2(t+8)

We multiply parentheses

2t+16

Back to the equation:

-(2t+16)


We get rid of parentheses

9t^2-12t-2t-16=0

We add all the numbers together, and all the variables

9t^2-14t-16=0

a = 9; b = -14; c = -16;
Δ = b2-4ac
Δ = -142-4·9·(-16)
Δ = 772
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:

$\sqrt{\Delta}=\sqrt{772}=\sqrt{4*193}=\sqrt{4}*\sqrt{193}=2\sqrt{193}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{193}}{2*9}=\frac{14-2\sqrt{193}}{18}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{193}}{2*9}=\frac{14+2\sqrt{193}}{18}
$