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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+6x^2-10x+3x-5)-3=0

We get rid of parentheses

6x^2-10x+3x-5-3=0

We add all the numbers together, and all the variables

6x^2-7x-8=0

a = 6; b = -7; c = -8;
Δ = b2-4ac
Δ = -72-4·6·(-8)
Δ = 241
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{-(-7)-\sqrt{241}}{2*6}=\frac{7-\sqrt{241}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{241}}{2*6}=\frac{7+\sqrt{241}}{12}
$