(4t+8)(2t-9)=0

Solution for (4t+8)(2t-9)=0 equation:

(4t+8)(2t-9)=0

We multiply parentheses ..

(+8t^2-36t+16t-72)=0

We get rid of parentheses

8t^2-36t+16t-72=0

We add all the numbers together, and all the variables

8t^2-20t-72=0

a = 8; b = -20; c = -72;
Δ = b2-4ac
Δ = -202-4·8·(-72)
Δ = 2704
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2704}=52$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-52}{2*8}=\frac{-32}{16}
=-2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+52}{2*8}=\frac{72}{16}
=4+1/2
$