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

Solution for 0=(8t+4)(-2t+10) equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

-((-16t^2+80t-8t+40))=0


We calculate terms in parentheses: -((-16t^2+80t-8t+40)), so:

(-16t^2+80t-8t+40)

We get rid of parentheses

-16t^2+80t-8t+40

We add all the numbers together, and all the variables

-16t^2+72t+40

Back to the equation:

-(-16t^2+72t+40)


We get rid of parentheses

16t^2-72t-40=0

a = 16; b = -72; c = -40;
Δ = b2-4ac
Δ = -722-4·16·(-40)
Δ = 7744
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{7744}=88$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-72)-88}{2*16}=\frac{-16}{32}
=-1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-72)+88}{2*16}=\frac{160}{32}
=5
$