1xx(3x+1)=24

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

1xx(3x+1)=24

We move all terms to the left:

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

We multiply parentheses

3x^2+x-24=0

a = 3; b = 1; c = -24;
Δ = b2-4ac
Δ = 12-4·3·(-24)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-17}{2*3}=\frac{-18}{6}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+17}{2*3}=\frac{16}{6}
=2+2/3
$