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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

4(-1x+3)-(-2(-1x+2)x)+8=0

We multiply parentheses

-4x-(-2(-1x+2)x)+12+8=0


We calculate terms in parentheses: -(-2(-1x+2)x), so:

-2(-1x+2)x

We multiply parentheses

2x^2-4x

Back to the equation:

-(2x^2-4x)


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-2x^2+20=0

a = -2; b = 0; c = +20;
Δ = b2-4ac
Δ = 02-4·(-2)·20
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{10}}{2*-2}=\frac{0-4\sqrt{10}}{-4}
=-\frac{4\sqrt{10}}{-4}
=-\frac{\sqrt{10}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{10}}{2*-2}=\frac{0+4\sqrt{10}}{-4}
=\frac{4\sqrt{10}}{-4}
=\frac{\sqrt{10}}{-1}
$