2(x^2-4x+4)=x+2x+3

Solution for 2(x^2-4x+4)=x+2x+3 equation:

2(x^2-4x+4)=x+2x+3

We move all terms to the left:

2(x^2-4x+4)-(x+2x+3)=0

We add all the numbers together, and all the variables

2(x^2-4x+4)-(3x+3)=0

We multiply parentheses

2x^2-8x-(3x+3)+8=0

We get rid of parentheses

2x^2-8x-3x-3+8=0

We add all the numbers together, and all the variables

2x^2-11x+5=0

a = 2; b = -11; c = +5;
Δ = b2-4ac
Δ = -112-4·2·5
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-9}{2*2}=\frac{2}{4}
=1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+9}{2*2}=\frac{20}{4}
=5
$