4(3+x)+-4=8x+10+-4x(3*4+x*4)+-4=8x+10+-4x(12+4x)+-4=8x+10+-4x

Solution for 4(3+x)+-4=8x+10+-4x(3*4+x*4)+-4=8x+10+-4x(12+4x)+-4=8x+10+-4x equation:

4(3+x)+-4=8x+10+-4x(3*4+x*4)+-4=8x+10+-4x(12+4x)+-4=8x+10+-4x

We move all terms to the left:

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

We add all the numbers together, and all the variables

4(x+3)-(8x+10+-4x(x*4+12)+-4)-4+=0

We add all the numbers together, and all the variables

4(x+3)-(8x+10+-4x(x*4+12)+-4)=0

We use the square of the difference formula

4(x+3)-(8x+10-4x(x*4+12)-4)=0

We multiply parentheses

4x-(8x+10-4x(x*4+12)-4)+12=0


We calculate terms in parentheses: -(8x+10-4x(x*4+12)-4), so:

8x+10-4x(x*4+12)-4

determiningTheFunctionDomain
8x-4x(x*4+12)+10-4

We add all the numbers together, and all the variables

8x-4x(x*4+12)+6

We multiply parentheses

-16x^2+8x-48x+6

We add all the numbers together, and all the variables

-16x^2-40x+6

Back to the equation:

-(-16x^2-40x+6)


We get rid of parentheses

16x^2+40x+4x-6+12=0

We add all the numbers together, and all the variables

16x^2+44x+6=0

a = 16; b = 44; c = +6;
Δ = b2-4ac
Δ = 442-4·16·6
Δ = 1552
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:

$\sqrt{\Delta}=\sqrt{1552}=\sqrt{16*97}=\sqrt{16}*\sqrt{97}=4\sqrt{97}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(44)-4\sqrt{97}}{2*16}=\frac{-44-4\sqrt{97}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(44)+4\sqrt{97}}{2*16}=\frac{-44+4\sqrt{97}}{32}
$