2(3x+2-8)x+4=2(2x-10)2x+3

Solution for 2(3x+2-8)x+4=2(2x-10)2x+3 equation:

2(3x+2-8)x+4=2(2x-10)2x+3

We move all terms to the left:

2(3x+2-8)x+4-(2(2x-10)2x+3)=0

We add all the numbers together, and all the variables

2(3x-6)x-(2(2x-10)2x+3)+4=0

We multiply parentheses

6x^2-12x-(2(2x-10)2x+3)+4=0


We calculate terms in parentheses: -(2(2x-10)2x+3), so:

2(2x-10)2x+3

We multiply parentheses

8x^2-40x+3

Back to the equation:

-(8x^2-40x+3)


We get rid of parentheses

6x^2-8x^2-12x+40x-3+4=0

We add all the numbers together, and all the variables

-2x^2+28x+1=0

a = -2; b = 28; c = +1;
Δ = b2-4ac
Δ = 282-4·(-2)·1
Δ = 792
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{792}=\sqrt{36*22}=\sqrt{36}*\sqrt{22}=6\sqrt{22}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-6\sqrt{22}}{2*-2}=\frac{-28-6\sqrt{22}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+6\sqrt{22}}{2*-2}=\frac{-28+6\sqrt{22}}{-4}
$