(3x+1)(x+5)=48

Solution for (3x+1)(x+5)=48 equation:

(3x+1)(x+5)=48

We move all terms to the left:

(3x+1)(x+5)-(48)=0

We multiply parentheses ..

(+3x^2+15x+x+5)-48=0

We get rid of parentheses

3x^2+15x+x+5-48=0

We add all the numbers together, and all the variables

3x^2+16x-43=0

a = 3; b = 16; c = -43;
Δ = b2-4ac
Δ = 162-4·3·(-43)
Δ = 772
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:

$\sqrt{\Delta}=\sqrt{772}=\sqrt{4*193}=\sqrt{4}*\sqrt{193}=2\sqrt{193}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{193}}{2*3}=\frac{-16-2\sqrt{193}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{193}}{2*3}=\frac{-16+2\sqrt{193}}{6}
$