0=-4(4t^2-17t-12)

Solution for 0=-4(4t^2-17t-12) equation:

0=-4(4t^2-17t-12)

We move all terms to the left:

0-(-4(4t^2-17t-12))=0

We add all the numbers together, and all the variables

-(-4(4t^2-17t-12))=0


We calculate terms in parentheses: -(-4(4t^2-17t-12)), so:

-4(4t^2-17t-12)

We multiply parentheses

-16t^2+68t+48

Back to the equation:

-(-16t^2+68t+48)


We get rid of parentheses

16t^2-68t-48=0

a = 16; b = -68; c = -48;
Δ = b2-4ac
Δ = -682-4·16·(-48)
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{-(-68)-4\sqrt{481}}{2*16}=\frac{68-4\sqrt{481}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-68)+4\sqrt{481}}{2*16}=\frac{68+4\sqrt{481}}{32}
$