(8t+16)(-3t+12)=6t-42+8

Solution for (8t+16)(-3t+12)=6t-42+8 equation:

(8t+16)(-3t+12)=6t-42+8

We move all terms to the left:

(8t+16)(-3t+12)-(6t-42+8)=0

We add all the numbers together, and all the variables

(8t+16)(-3t+12)-(6t-34)=0

We get rid of parentheses

(8t+16)(-3t+12)-6t+34=0

We multiply parentheses ..

(-24t^2+96t-48t+192)-6t+34=0

We get rid of parentheses

-24t^2+96t-48t-6t+192+34=0

We add all the numbers together, and all the variables

-24t^2+42t+226=0

a = -24; b = 42; c = +226;
Δ = b2-4ac
Δ = 422-4·(-24)·226
Δ = 23460
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:

$\sqrt{\Delta}=\sqrt{23460}=\sqrt{4*5865}=\sqrt{4}*\sqrt{5865}=2\sqrt{5865}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-2\sqrt{5865}}{2*-24}=\frac{-42-2\sqrt{5865}}{-48}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+2\sqrt{5865}}{2*-24}=\frac{-42+2\sqrt{5865}}{-48}
$