2x+(-3/4x+9)=16

Solution for 2x+(-3/4x+9)=16 equation:

2x+(-3/4x+9)=16

We move all terms to the left:

2x+(-3/4x+9)-(16)=0


Domain of the equation: 4x+9)!=0

x∈R


We get rid of parentheses

2x-3/4x+9-16=0

We multiply all the terms by the denominator


2x*4x+9*4x-16*4x-3=0

Wy multiply elements

8x^2+36x-64x-3=0

We add all the numbers together, and all the variables

8x^2-28x-3=0

a = 8; b = -28; c = -3;
Δ = b2-4ac
Δ = -282-4·8·(-3)
Δ = 880
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:

$\sqrt{\Delta}=\sqrt{880}=\sqrt{16*55}=\sqrt{16}*\sqrt{55}=4\sqrt{55}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-4\sqrt{55}}{2*8}=\frac{28-4\sqrt{55}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+4\sqrt{55}}{2*8}=\frac{28+4\sqrt{55}}{16}
$