0=-4(4t+5)(t-3)

Solution for 0=-4(4t+5)(t-3) equation:

0=-4(4t+5)(t-3)

We move all terms to the left:

0-(-4(4t+5)(t-3))=0

We add all the numbers together, and all the variables

-(-4(4t+5)(t-3))=0

We multiply parentheses ..

-(-4(+4t^2-12t+5t-15))=0


We calculate terms in parentheses: -(-4(+4t^2-12t+5t-15)), so:

-4(+4t^2-12t+5t-15)

We multiply parentheses

-16t^2+48t-20t+60

We add all the numbers together, and all the variables

-16t^2+28t+60

Back to the equation:

-(-16t^2+28t+60)


We get rid of parentheses

16t^2-28t-60=0

a = 16; b = -28; c = -60;
Δ = b2-4ac
Δ = -282-4·16·(-60)
Δ = 4624
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4624}=68$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-68}{2*16}=\frac{-40}{32}
=-1+1/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+68}{2*16}=\frac{96}{32}
=3
$