2(3+x)+x(1-4x)+5=7x+4x+5

Solution for 2(3+x)+x(1-4x)+5=7x+4x+5 equation:

2(3+x)+x(1-4x)+5=7x+4x+5

We move all terms to the left:

2(3+x)+x(1-4x)+5-(7x+4x+5)=0

We add all the numbers together, and all the variables

2(x+3)+x(-4x+1)-(11x+5)+5=0

We multiply parentheses

-4x^2+2x+x-(11x+5)+6+5=0

We get rid of parentheses

-4x^2+2x+x-11x-5+6+5=0

We add all the numbers together, and all the variables

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

a = -4; b = -8; c = +6;
Δ = b2-4ac
Δ = -82-4·(-4)·6
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{-(-8)-4\sqrt{10}}{2*-4}=\frac{8-4\sqrt{10}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{10}}{2*-4}=\frac{8+4\sqrt{10}}{-8}
$