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

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

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

We move all terms to the left:

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

We get rid of parentheses

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


We calculate terms in parentheses: -((3x^2-2x-5)), so:

(3x^2-2x-5)

We get rid of parentheses

3x^2-2x-5

Back to the equation:

-(3x^2-2x-5)


We get rid of parentheses

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

We add all the numbers together, and all the variables

-3x^2+4x+6=0

a = -3; b = 4; c = +6;
Δ = b2-4ac
Δ = 42-4·(-3)·6
Δ = 88
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:

$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{22}}{2*-3}=\frac{-4-2\sqrt{22}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{22}}{2*-3}=\frac{-4+2\sqrt{22}}{-6}
$