3x+(5x+x)=x+14-(7x2)

Solution for 3x+(5x+x)=x+14-(7x2) equation:

3x+(5x+x)=x+14-(7x^2)

We move all terms to the left:

3x+(5x+x)-(x+14-(7x^2))=0

determiningTheFunctionDomain
-(x+14-7x^2)+3x+(5x+x)=0

We add all the numbers together, and all the variables

-(x+14-7x^2)+3x+(+6x)=0

We get rid of parentheses

7x^2-x+3x+6x-14=0

We add all the numbers together, and all the variables

7x^2+8x-14=0

a = 7; b = 8; c = -14;
Δ = b2-4ac
Δ = 82-4·7·(-14)
Δ = 456
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:

$\sqrt{\Delta}=\sqrt{456}=\sqrt{4*114}=\sqrt{4}*\sqrt{114}=2\sqrt{114}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{114}}{2*7}=\frac{-8-2\sqrt{114}}{14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{114}}{2*7}=\frac{-8+2\sqrt{114}}{14}
$