(3x+1)(x+4)=22

Solution for (3x+1)(x+4)=22 equation:

(3x+1)(x+4)=22

We move all terms to the left:

(3x+1)(x+4)-(22)=0

We multiply parentheses ..

(+3x^2+12x+x+4)-22=0

We get rid of parentheses

3x^2+12x+x+4-22=0

We add all the numbers together, and all the variables

3x^2+13x-18=0

a = 3; b = 13; c = -18;
Δ = b2-4ac
Δ = 132-4·3·(-18)
Δ = 385
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{-(13)-\sqrt{385}}{2*3}=\frac{-13-\sqrt{385}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{385}}{2*3}=\frac{-13+\sqrt{385}}{6}
$