(3x2-5x+1)=(x2+4x+2)

Solution for (3x2-5x+1)=(x2+4x+2) equation:

(3x^2-5x+1)=(x2+4x+2)

We move all terms to the left:

(3x^2-5x+1)-((x2+4x+2))=0

We add all the numbers together, and all the variables

-((+x^2+4x+2))+(3x^2-5x+1)=0

We get rid of parentheses

-((+x^2+4x+2))+3x^2-5x+1=0


We calculate terms in parentheses: -((+x^2+4x+2)), so:

(+x^2+4x+2)

We get rid of parentheses

x^2+4x+2

Back to the equation:

-(x^2+4x+2)


We add all the numbers together, and all the variables

3x^2-5x-(x^2+4x+2)+1=0

We get rid of parentheses

3x^2-x^2-5x-4x-2+1=0

We add all the numbers together, and all the variables

2x^2-9x-1=0

a = 2; b = -9; c = -1;
Δ = b2-4ac
Δ = -92-4·2·(-1)
Δ = 89
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{-(-9)-\sqrt{89}}{2*2}=\frac{9-\sqrt{89}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{89}}{2*2}=\frac{9+\sqrt{89}}{4}
$