(2x+1)(3x-5)+1=4x^2

Solution for (2x+1)(3x-5)+1=4x^2 equation:

(2x+1)(3x-5)+1=4x^2

We move all terms to the left:

(2x+1)(3x-5)+1-(4x^2)=0

determiningTheFunctionDomain
-4x^2+(2x+1)(3x-5)+1=0

We multiply parentheses ..

-4x^2+(+6x^2-10x+3x-5)+1=0

We get rid of parentheses

-4x^2+6x^2-10x+3x-5+1=0

We add all the numbers together, and all the variables

2x^2-7x-4=0

a = 2; b = -7; c = -4;
Δ = b2-4ac
Δ = -72-4·2·(-4)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-9}{2*2}=\frac{-2}{4}
=-1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+9}{2*2}=\frac{16}{4}
=4
$