2x+(3/4)(4x+16)=7

Solution for 2x+(3/4)(4x+16)=7 equation:

2x+(3/4)(4x+16)=7

We move all terms to the left:

2x+(3/4)(4x+16)-(7)=0


Domain of the equation: 4)(4x+16)!=0

x∈R


We add all the numbers together, and all the variables

2x+(+3/4)(4x+16)-7=0

We multiply parentheses ..

(+12x^2+3/4*16)+2x-7=0

We multiply all the terms by the denominator


(+12x^2+3+2x*4*16)-7*4*16)=0

We add all the numbers together, and all the variables

(+12x^2+3+2x*4*16)=0

We get rid of parentheses

12x^2+2x*4*16+3=0

Wy multiply elements

12x^2+128x*1+3=0

Wy multiply elements

12x^2+128x+3=0

a = 12; b = 128; c = +3;
Δ = b2-4ac
Δ = 1282-4·12·3
Δ = 16240
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:

$\sqrt{\Delta}=\sqrt{16240}=\sqrt{16*1015}=\sqrt{16}*\sqrt{1015}=4\sqrt{1015}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(128)-4\sqrt{1015}}{2*12}=\frac{-128-4\sqrt{1015}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(128)+4\sqrt{1015}}{2*12}=\frac{-128+4\sqrt{1015}}{24}
$