-3(t-4)6t=8t-5

Solution for -3(t-4)6t=8t-5 equation:

-3(t-4)6t=8t-5

We move all terms to the left:

-3(t-4)6t-(8t-5)=0

We multiply parentheses

-18t^2+72t-(8t-5)=0

We get rid of parentheses

-18t^2+72t-8t+5=0

We add all the numbers together, and all the variables

-18t^2+64t+5=0

a = -18; b = 64; c = +5;
Δ = b2-4ac
Δ = 642-4·(-18)·5
Δ = 4456
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:

$\sqrt{\Delta}=\sqrt{4456}=\sqrt{4*1114}=\sqrt{4}*\sqrt{1114}=2\sqrt{1114}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-2\sqrt{1114}}{2*-18}=\frac{-64-2\sqrt{1114}}{-36}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+2\sqrt{1114}}{2*-18}=\frac{-64+2\sqrt{1114}}{-36}
$