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

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

=(2E+4)(E-2)+(E+2)(4E-7)

We move all terms to the left:

-((2E+4)(E-2)+(E+2)(4E-7))=0

We multiply parentheses ..

-((+2E^2-4E+4E-8)+(E+2)(4E-7))=0


We calculate terms in parentheses: -((+2E^2-4E+4E-8)+(E+2)(4E-7)), so:

(+2E^2-4E+4E-8)+(E+2)(4E-7)

We get rid of parentheses

2E^2-4E+4E+(E+2)(4E-7)-8

We multiply parentheses ..

2E^2+(+4E^2-7E+8E-14)-4E+4E-8

We add all the numbers together, and all the variables

2E^2+(+4E^2-7E+8E-14)-8

We get rid of parentheses

2E^2+4E^2-7E+8E-14-8

We add all the numbers together, and all the variables

6E^2+E-22

Back to the equation:

-(6E^2+E-22)


We get rid of parentheses

-6E^2-E+22=0

We add all the numbers together, and all the variables

-6E^2-1E+22=0

a = -6; b = -1; c = +22;
Δ = b2-4ac
Δ = -12-4·(-6)·22
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$E_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$E_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$E_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-23}{2*-6}=\frac{-22}{-12}
=1+5/6
$
$E_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+23}{2*-6}=\frac{24}{-12}
=-2
$