(6-5t)(2t+6)=4

Solution for (6-5t)(2t+6)=4 equation:

(6-5t)(2t+6)=4

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-10t^2-30t+12t+36)-4=0

We get rid of parentheses

-10t^2-30t+12t+36-4=0

We add all the numbers together, and all the variables

-10t^2-18t+32=0

a = -10; b = -18; c = +32;
Δ = b2-4ac
Δ = -182-4·(-10)·32
Δ = 1604
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:

$\sqrt{\Delta}=\sqrt{1604}=\sqrt{4*401}=\sqrt{4}*\sqrt{401}=2\sqrt{401}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{401}}{2*-10}=\frac{18-2\sqrt{401}}{-20}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{401}}{2*-10}=\frac{18+2\sqrt{401}}{-20}
$