4x(2x+1)=5x+(3x+9)

Solution for 4x(2x+1)=5x+(3x+9) equation:

4x(2x+1)=5x+(3x+9)

We move all terms to the left:

4x(2x+1)-(5x+(3x+9))=0

We multiply parentheses

8x^2+4x-(5x+(3x+9))=0


We calculate terms in parentheses: -(5x+(3x+9)), so:

5x+(3x+9)

We get rid of parentheses

5x+3x+9

We add all the numbers together, and all the variables

8x+9

Back to the equation:

-(8x+9)


We get rid of parentheses

8x^2+4x-8x-9=0

We add all the numbers together, and all the variables

8x^2-4x-9=0

a = 8; b = -4; c = -9;
Δ = b2-4ac
Δ = -42-4·8·(-9)
Δ = 304
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:

$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{19}}{2*8}=\frac{4-4\sqrt{19}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{19}}{2*8}=\frac{4+4\sqrt{19}}{16}
$