(2t+8)(3-t)=0

Solution for (2t+8)(3-t)=0 equation:

(2t+8)(3-t)=0

We add all the numbers together, and all the variables

(2t+8)(-1t+3)=0

We multiply parentheses ..

(-2t^2+6t-8t+24)=0

We get rid of parentheses

-2t^2+6t-8t+24=0

We add all the numbers together, and all the variables

-2t^2-2t+24=0

a = -2; b = -2; c = +24;
Δ = b2-4ac
Δ = -22-4·(-2)·24
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-14}{2*-2}=\frac{-12}{-4}
=+3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+14}{2*-2}=\frac{16}{-4}
=-4
$