1=x^2/3x+4

Solution for 1=x^2/3x+4 equation:

1=x^2/3x+4

We move all terms to the left:

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


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

x∈R


We get rid of parentheses

-x^2/3x-4+1=0

We multiply all the terms by the denominator


-x^2-4*3x+1*3x=0

We add all the numbers together, and all the variables

-1x^2-4*3x+1*3x=0

Wy multiply elements

-1x^2-12x+3x=0

We add all the numbers together, and all the variables

-1x^2-9x=0

a = -1; b = -9; c = 0;
Δ = b2-4ac
Δ = -92-4·(-1)·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-9}{2*-1}=\frac{0}{-2}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+9}{2*-1}=\frac{18}{-2}
=-9
$