3x+4=8x(x+1)

Solution for 3x+4=8x(x+1) equation:

3x+4=8x(x+1)

We move all terms to the left:

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


We calculate terms in parentheses: -(8x(x+1)), so:

8x(x+1)

We multiply parentheses

8x^2+8x

Back to the equation:

-(8x^2+8x)


We get rid of parentheses

-8x^2+3x-8x+4=0

We add all the numbers together, and all the variables

-8x^2-5x+4=0

a = -8; b = -5; c = +4;
Δ = b2-4ac
Δ = -52-4·(-8)·4
Δ = 153
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:

$\sqrt{\Delta}=\sqrt{153}=\sqrt{9*17}=\sqrt{9}*\sqrt{17}=3\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-3\sqrt{17}}{2*-8}=\frac{5-3\sqrt{17}}{-16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+3\sqrt{17}}{2*-8}=\frac{5+3\sqrt{17}}{-16}
$