4t=-(4t-12)(4t-36)

Solution for 4t=-(4t-12)(4t-36) equation:

4t=-(4t-12)(4t-36)

We move all terms to the left:

4t-(-(4t-12)(4t-36))=0

We multiply parentheses ..

-(-(+16t^2-144t-48t+432))+4t=0


We calculate terms in parentheses: -(-(+16t^2-144t-48t+432)), so:

-(+16t^2-144t-48t+432)

We get rid of parentheses

-16t^2+144t+48t-432

We add all the numbers together, and all the variables

-16t^2+192t-432

Back to the equation:

-(-16t^2+192t-432)


We get rid of parentheses

16t^2-192t+4t+432=0

We add all the numbers together, and all the variables

16t^2-188t+432=0

a = 16; b = -188; c = +432;
Δ = b2-4ac
Δ = -1882-4·16·432
Δ = 7696
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:

$\sqrt{\Delta}=\sqrt{7696}=\sqrt{16*481}=\sqrt{16}*\sqrt{481}=4\sqrt{481}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-188)-4\sqrt{481}}{2*16}=\frac{188-4\sqrt{481}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-188)+4\sqrt{481}}{2*16}=\frac{188+4\sqrt{481}}{32}
$