42=(x+3)(2x+8)

Solution for 42=(x+3)(2x+8) equation:

42=(x+3)(2x+8)

We move all terms to the left:

42-((x+3)(2x+8))=0

We multiply parentheses ..

-((+2x^2+8x+6x+24))+42=0


We calculate terms in parentheses: -((+2x^2+8x+6x+24)), so:

(+2x^2+8x+6x+24)

We get rid of parentheses

2x^2+8x+6x+24

We add all the numbers together, and all the variables

2x^2+14x+24

Back to the equation:

-(2x^2+14x+24)


We get rid of parentheses

-2x^2-14x-24+42=0

We add all the numbers together, and all the variables

-2x^2-14x+18=0

a = -2; b = -14; c = +18;
Δ = b2-4ac
Δ = -142-4·(-2)·18
Δ = 340
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{340}=\sqrt{4*85}=\sqrt{4}*\sqrt{85}=2\sqrt{85}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{85}}{2*-2}=\frac{14-2\sqrt{85}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{85}}{2*-2}=\frac{14+2\sqrt{85}}{-4}
$