(5x+2)2-(7x+3)(3x-2)=4x2

Solution for (5x+2)2-(7x+3)(3x-2)=4x2 equation:

(5x+2)2-(7x+3)(3x-2)=4x^2

We move all terms to the left:

(5x+2)2-(7x+3)(3x-2)-(4x^2)=0

determiningTheFunctionDomain
-4x^2+(5x+2)2-(7x+3)(3x-2)=0

We multiply parentheses

-4x^2+10x-(7x+3)(3x-2)+4=0

We multiply parentheses ..

-4x^2-(+21x^2-14x+9x-6)+10x+4=0

We get rid of parentheses

-4x^2-21x^2+14x-9x+10x+6+4=0

We add all the numbers together, and all the variables

-25x^2+15x+10=0

a = -25; b = 15; c = +10;
Δ = b2-4ac
Δ = 152-4·(-25)·10
Δ = 1225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1225}=35$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-35}{2*-25}=\frac{-50}{-50}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+35}{2*-25}=\frac{20}{-50}
=-2/5
$