(3t-2)(t+2)-3t*2=5

Solution for (3t-2)(t+2)-3t*2=5 equation:

(3t-2)(t+2)-3t*2=5

We move all terms to the left:

(3t-2)(t+2)-3t*2-(5)=0

Wy multiply elements

(3t-2)(t+2)-6t-5=0

We multiply parentheses ..

(+3t^2+6t-2t-4)-6t-5=0

We get rid of parentheses

3t^2+6t-2t-6t-4-5=0

We add all the numbers together, and all the variables

3t^2-2t-9=0

a = 3; b = -2; c = -9;
Δ = b2-4ac
Δ = -22-4·3·(-9)
Δ = 112
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:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-4\sqrt{7}}{2*3}=\frac{2-4\sqrt{7}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+4\sqrt{7}}{2*3}=\frac{2+4\sqrt{7}}{6}
$