24=(x+1)(3x+1)

Solution for 24=(x+1)(3x+1) equation:

24=(x+1)(3x+1)

We move all terms to the left:

24-((x+1)(3x+1))=0

We multiply parentheses ..

-((+3x^2+x+3x+1))+24=0


We calculate terms in parentheses: -((+3x^2+x+3x+1)), so:

(+3x^2+x+3x+1)

We get rid of parentheses

3x^2+x+3x+1

We add all the numbers together, and all the variables

3x^2+4x+1

Back to the equation:

-(3x^2+4x+1)


We get rid of parentheses

-3x^2-4x-1+24=0

We add all the numbers together, and all the variables

-3x^2-4x+23=0

a = -3; b = -4; c = +23;
Δ = b2-4ac
Δ = -42-4·(-3)·23
Δ = 292
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:

$\sqrt{\Delta}=\sqrt{292}=\sqrt{4*73}=\sqrt{4}*\sqrt{73}=2\sqrt{73}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{73}}{2*-3}=\frac{4-2\sqrt{73}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{73}}{2*-3}=\frac{4+2\sqrt{73}}{-6}
$