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

Solution for (2x+1)(3x-5)-5=0 equation:

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

6x^2-7x-10=0

a = 6; b = -7; c = -10;
Δ = b2-4ac
Δ = -72-4·6·(-10)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-17}{2*6}=\frac{-10}{12}
=-5/6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+17}{2*6}=\frac{24}{12}
=2
$