(3/4)(2x+4)=x+19

Solution for (3/4)(2x+4)=x+19 equation:

(3/4)(2x+4)=x+19

We move all terms to the left:

(3/4)(2x+4)-(x+19)=0


Domain of the equation: 4)(2x+4)!=0

x∈R


We add all the numbers together, and all the variables

(+3/4)(2x+4)-(x+19)=0

We get rid of parentheses

(+3/4)(2x+4)-x-19=0

We multiply parentheses ..

(+6x^2+3/4*4)-x-19=0

We multiply all the terms by the denominator


(+6x^2+3-x*4*4)-19*4*4)=0

We add all the numbers together, and all the variables

(+6x^2+3-x*4*4)=0

We get rid of parentheses

6x^2-x*4*4+3=0

Wy multiply elements

6x^2-16x*4+3=0

Wy multiply elements

6x^2-64x+3=0

a = 6; b = -64; c = +3;
Δ = b2-4ac
Δ = -642-4·6·3
Δ = 4024
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:

$\sqrt{\Delta}=\sqrt{4024}=\sqrt{4*1006}=\sqrt{4}*\sqrt{1006}=2\sqrt{1006}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-64)-2\sqrt{1006}}{2*6}=\frac{64-2\sqrt{1006}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-64)+2\sqrt{1006}}{2*6}=\frac{64+2\sqrt{1006}}{12}
$