3x2-7=4/7+x

Solution for 3x2-7=4/7+x equation:

3x^2-7=4/7+x

We move all terms to the left:

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


Domain of the equation: 7+x)!=0

We move all terms containing x to the left, all other terms to the right

x)!=-7

x!=-7/1

x!=-7

x∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

3x^2-x-7-4/7=0

We multiply all the terms by the denominator


3x^2*7-x*7-4-7*7=0

We add all the numbers together, and all the variables

3x^2*7-x*7-53=0

Wy multiply elements

21x^2-7x-53=0

a = 21; b = -7; c = -53;
Δ = b2-4ac
Δ = -72-4·21·(-53)
Δ = 4501
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-\sqrt{4501}}{2*21}=\frac{7-\sqrt{4501}}{42}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{4501}}{2*21}=\frac{7+\sqrt{4501}}{42}
$