2(x^2+2x+4)-(x-2)(x-3)=2(4x+5)

Solution for 2(x^2+2x+4)-(x-2)(x-3)=2(4x+5) equation:

2(x^2+2x+4)-(x-2)(x-3)=2(4x+5)

We move all terms to the left:

2(x^2+2x+4)-(x-2)(x-3)-(2(4x+5))=0

We multiply parentheses

2x^2+4x-(x-2)(x-3)-(2(4x+5))+8=0

We multiply parentheses ..

2x^2-(+x^2-3x-2x+6)+4x-(2(4x+5))+8=0


We calculate terms in parentheses: -(2(4x+5)), so:

2(4x+5)

We multiply parentheses

8x+10

Back to the equation:

-(8x+10)


We get rid of parentheses

2x^2-x^2+3x+2x+4x-8x-6-10+8=0

We add all the numbers together, and all the variables

x^2+x-8=0

a = 1; b = 1; c = -8;
Δ = b2-4ac
Δ = 12-4·1·(-8)
Δ = 33
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{33}}{2*1}=\frac{-1-\sqrt{33}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{33}}{2*1}=\frac{-1+\sqrt{33}}{2}
$