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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

6x^2-13x-10=0

a = 6; b = -13; c = -10;
Δ = b2-4ac
Δ = -132-4·6·(-10)
Δ = 409
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{409}}{2*6}=\frac{13-\sqrt{409}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{409}}{2*6}=\frac{13+\sqrt{409}}{12}
$