4x(2x-4)=7x(1x+5)

Solution for 4x(2x-4)=7x(1x+5) equation:

4x(2x-4)=7x(1x+5)

We move all terms to the left:

4x(2x-4)-(7x(1x+5))=0

We add all the numbers together, and all the variables

4x(2x-4)-(7x(x+5))=0

We multiply parentheses

8x^2-16x-(7x(x+5))=0


We calculate terms in parentheses: -(7x(x+5)), so:

7x(x+5)

We multiply parentheses

7x^2+35x

Back to the equation:

-(7x^2+35x)


We get rid of parentheses

8x^2-7x^2-16x-35x=0

We add all the numbers together, and all the variables

x^2-51x=0

a = 1; b = -51; c = 0;
Δ = b2-4ac
Δ = -512-4·1·0
Δ = 2601
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2601}=51$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-51)-51}{2*1}=\frac{0}{2}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-51)+51}{2*1}=\frac{102}{2}
=51
$