(t+8)(2t-7)=0

Solution for (t+8)(2t-7)=0 equation:

(t+8)(2t-7)=0

We multiply parentheses ..

(+2t^2-7t+16t-56)=0

We get rid of parentheses

2t^2-7t+16t-56=0

We add all the numbers together, and all the variables

2t^2+9t-56=0

a = 2; b = 9; c = -56;
Δ = b2-4ac
Δ = 92-4·2·(-56)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-23}{2*2}=\frac{-32}{4}
=-8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+23}{2*2}=\frac{14}{4}
=3+1/2
$