4x^2+4+x(x-2)=10(x-2)+20

Solution for 4x^2+4+x(x-2)=10(x-2)+20 equation:

4x^2+4+x(x-2)=10(x-2)+20

We move all terms to the left:

4x^2+4+x(x-2)-(10(x-2)+20)=0

We multiply parentheses

4x^2+x^2-2x-(10(x-2)+20)+4=0


We calculate terms in parentheses: -(10(x-2)+20), so:

10(x-2)+20

We multiply parentheses

10x-20+20

We add all the numbers together, and all the variables

10x

Back to the equation:

-(10x)


We add all the numbers together, and all the variables

5x^2-12x+4=0

a = 5; b = -12; c = +4;
Δ = b2-4ac
Δ = -122-4·5·4
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-8}{2*5}=\frac{4}{10}
=2/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+8}{2*5}=\frac{20}{10}
=2
$