5=(7/2)(3x+)

Solution for 5=(7/2)(3x+) equation:

5=(7/2)(3x+)

We move all terms to the left:

5-((7/2)(3x+))=0


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

x∈R


We add all the numbers together, and all the variables

-((+7/2)(+3x))+5=0

We multiply parentheses ..

-((+21x^2))+5=0


We calculate terms in parentheses: -((+21x^2)), so:

(+21x^2)

We get rid of parentheses

21x^2

Back to the equation:

-(21x^2)


a = -21; b = 0; c = +5;
Δ = b2-4ac
Δ = 02-4·(-21)·5
Δ = 420
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:

$\sqrt{\Delta}=\sqrt{420}=\sqrt{4*105}=\sqrt{4}*\sqrt{105}=2\sqrt{105}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{105}}{2*-21}=\frac{0-2\sqrt{105}}{-42}
=-\frac{2\sqrt{105}}{-42}
=-\frac{\sqrt{105}}{-21}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{105}}{2*-21}=\frac{0+2\sqrt{105}}{-42}
=\frac{2\sqrt{105}}{-42}
=\frac{\sqrt{105}}{-21}
$