7x(2x+5)-5x(2x+3)=(2x+4)23

Solution for 7x(2x+5)-5x(2x+3)=(2x+4)23 equation:

7x(2x+5)-5x(2x+3)=(2x+4)23

We move all terms to the left:

7x(2x+5)-5x(2x+3)-((2x+4)23)=0

We multiply parentheses

14x^2-10x^2+35x-15x-((2x+4)23)=0


We calculate terms in parentheses: -((2x+4)23), so:

(2x+4)23

We multiply parentheses

46x+92

Back to the equation:

-(46x+92)


We add all the numbers together, and all the variables

4x^2+20x-(46x+92)=0

We get rid of parentheses

4x^2+20x-46x-92=0

We add all the numbers together, and all the variables

4x^2-26x-92=0

a = 4; b = -26; c = -92;
Δ = b2-4ac
Δ = -262-4·4·(-92)
Δ = 2148
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2148}=\sqrt{4*537}=\sqrt{4}*\sqrt{537}=2\sqrt{537}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-2\sqrt{537}}{2*4}=\frac{26-2\sqrt{537}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+2\sqrt{537}}{2*4}=\frac{26+2\sqrt{537}}{8}
$