0=8(-2t+5)(t+2)

Solution for 0=8(-2t+5)(t+2) equation:

0=8(-2t+5)(t+2)

We move all terms to the left:

0-(8(-2t+5)(t+2))=0

We add all the numbers together, and all the variables

-(8(-2t+5)(t+2))=0

We multiply parentheses ..

-(8(-2t^2-4t+5t+10))=0


We calculate terms in parentheses: -(8(-2t^2-4t+5t+10)), so:

8(-2t^2-4t+5t+10)

We multiply parentheses

-16t^2-32t+40t+80

We add all the numbers together, and all the variables

-16t^2+8t+80

Back to the equation:

-(-16t^2+8t+80)


We get rid of parentheses

16t^2-8t-80=0

a = 16; b = -8; c = -80;
Δ = b2-4ac
Δ = -82-4·16·(-80)
Δ = 5184
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5184}=72$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-72}{2*16}=\frac{-64}{32}
=-2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+72}{2*16}=\frac{80}{32}
=2+1/2
$