(3/4)*(3+2x)+1=13

Solution for (3/4)*(3+2x)+1=13 equation:

(3/4)(3+2x)+1=13

We move all terms to the left:

(3/4)(3+2x)+1-(13)=0


Domain of the equation: 4)(3+2x)!=0

x∈R


We add all the numbers together, and all the variables

(+3/4)(2x+3)+1-13=0

We add all the numbers together, and all the variables

(+3/4)(2x+3)-12=0

We multiply parentheses ..

(+6x^2+3/4*3)-12=0

We multiply all the terms by the denominator


(+6x^2+3-12*4*3)=0

We get rid of parentheses

6x^2+3-12*4*3=0

We add all the numbers together, and all the variables

6x^2-141=0

a = 6; b = 0; c = -141;
Δ = b2-4ac
Δ = 02-4·6·(-141)
Δ = 3384
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:

$\sqrt{\Delta}=\sqrt{3384}=\sqrt{36*94}=\sqrt{36}*\sqrt{94}=6\sqrt{94}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{94}}{2*6}=\frac{0-6\sqrt{94}}{12}
=-\frac{6\sqrt{94}}{12}
=-\frac{\sqrt{94}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{94}}{2*6}=\frac{0+6\sqrt{94}}{12}
=\frac{6\sqrt{94}}{12}
=\frac{\sqrt{94}}{2}
$