(-35)-7x=(-7x)(x+5)

Solution for (-35)-7x=(-7x)(x+5) equation:

(-35)-7x=(-7x)(x+5)

We move all terms to the left:

(-35)-7x-((-7x)(x+5))=0

We add all the numbers together, and all the variables

-7x-((-7x)(x+5))-35=0

We multiply parentheses ..

-((-7x^2-35x))-7x-35=0


We calculate terms in parentheses: -((-7x^2-35x)), so:

(-7x^2-35x)

We get rid of parentheses

-7x^2-35x

Back to the equation:

-(-7x^2-35x)


We get rid of parentheses

7x^2+35x-7x-35=0

We add all the numbers together, and all the variables

7x^2+28x-35=0

a = 7; b = 28; c = -35;
Δ = b2-4ac
Δ = 282-4·7·(-35)
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-42}{2*7}=\frac{-70}{14}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+42}{2*7}=\frac{14}{14}
=1
$