(3x+1)=(x2+2x-7)

Solution for (3x+1)=(x2+2x-7) equation:

(3x+1)=(x2+2x-7)

We move all terms to the left:

(3x+1)-((x2+2x-7))=0

We add all the numbers together, and all the variables

-((+x^2+2x-7))+(3x+1)=0

We get rid of parentheses

-((+x^2+2x-7))+3x+1=0


We calculate terms in parentheses: -((+x^2+2x-7)), so:

(+x^2+2x-7)

We get rid of parentheses

x^2+2x-7

Back to the equation:

-(x^2+2x-7)


We add all the numbers together, and all the variables

3x-(x^2+2x-7)+1=0

We get rid of parentheses

-x^2+3x-2x+7+1=0

We add all the numbers together, and all the variables

-1x^2+x+8=0

a = -1; b = 1; c = +8;
Δ = b2-4ac
Δ = 12-4·(-1)·8
Δ = 33
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{-(1)-\sqrt{33}}{2*-1}=\frac{-1-\sqrt{33}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{33}}{2*-1}=\frac{-1+\sqrt{33}}{-2}
$