42=(x+4)*(2x-7)

Solution for 42=(x+4)*(2x-7) equation:

42=(x+4)(2x-7)

We move all terms to the left:

42-((x+4)(2x-7))=0

We multiply parentheses ..

-((+2x^2-7x+8x-28))+42=0


We calculate terms in parentheses: -((+2x^2-7x+8x-28)), so:

(+2x^2-7x+8x-28)

We get rid of parentheses

2x^2-7x+8x-28

We add all the numbers together, and all the variables

2x^2+x-28

Back to the equation:

-(2x^2+x-28)


We get rid of parentheses

-2x^2-x+28+42=0

We add all the numbers together, and all the variables

-2x^2-1x+70=0

a = -2; b = -1; c = +70;
Δ = b2-4ac
Δ = -12-4·(-2)·70
Δ = 561
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{561}}{2*-2}=\frac{1-\sqrt{561}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{561}}{2*-2}=\frac{1+\sqrt{561}}{-4}
$