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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2j^2-9j-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:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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