(3x+1)*(2x+5)=8

Solution for (3x+1)*(2x+5)=8 equation:

(3x+1)(2x+5)=8

We move all terms to the left:

(3x+1)(2x+5)-(8)=0

We multiply parentheses ..

(+6x^2+15x+2x+5)-8=0

We get rid of parentheses

6x^2+15x+2x+5-8=0

We add all the numbers together, and all the variables

6x^2+17x-3=0

a = 6; b = 17; c = -3;
Δ = b2-4ac
Δ = 172-4·6·(-3)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-19}{2*6}=\frac{-36}{12}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+19}{2*6}=\frac{2}{12}
=1/6
$