4x^2-8x+4=4(x+5)

Solution for 4x^2-8x+4=4(x+5) equation:

4x^2-8x+4=4(x+5)

We move all terms to the left:

4x^2-8x+4-(4(x+5))=0


We calculate terms in parentheses: -(4(x+5)), so:

4(x+5)

We multiply parentheses

4x+20

Back to the equation:

-(4x+20)


We get rid of parentheses

4x^2-8x-4x-20+4=0

We add all the numbers together, and all the variables

4x^2-12x-16=0

a = 4; b = -12; c = -16;
Δ = b2-4ac
Δ = -122-4·4·(-16)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-20}{2*4}=\frac{-8}{8}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+20}{2*4}=\frac{32}{8}
=4
$