36=-4(4t^2-t-12)

Solution for 36=-4(4t^2-t-12) equation:

36=-4(4t^2-t-12)

We move all terms to the left:

36-(-4(4t^2-t-12))=0


We calculate terms in parentheses: -(-4(4t^2-t-12)), so:

-4(4t^2-t-12)

We multiply parentheses

-16t^2+4t+48

Back to the equation:

-(-16t^2+4t+48)


We get rid of parentheses

16t^2-4t-48+36=0

We add all the numbers together, and all the variables

16t^2-4t-12=0

a = 16; b = -4; c = -12;
Δ = b2-4ac
Δ = -42-4·16·(-12)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-28}{2*16}=\frac{-24}{32}
=-3/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+28}{2*16}=\frac{32}{32}
=1
$