(2x-42)(x+8)=0

Solution for (2x-42)(x+8)=0 equation:

(2x-42)(x+8)=0

We multiply parentheses ..

(+2x^2+16x-42x-336)=0

We get rid of parentheses

2x^2+16x-42x-336=0

We add all the numbers together, and all the variables

2x^2-26x-336=0

a = 2; b = -26; c = -336;
Δ = b2-4ac
Δ = -262-4·2·(-336)
Δ = 3364
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3364}=58$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-58}{2*2}=\frac{-32}{4}
=-8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+58}{2*2}=\frac{84}{4}
=21
$