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

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

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

We move all terms to the left:

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

We multiply parentheses

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


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

(2x+4)2

We multiply parentheses

4x+8

Back to the equation:

-(4x+8)


We add all the numbers together, and all the variables

4x^2+20x-(4x+8)=0

We get rid of parentheses

4x^2+20x-4x-8=0

We add all the numbers together, and all the variables

4x^2+16x-8=0

a = 4; b = 16; c = -8;
Δ = b2-4ac
Δ = 162-4·4·(-8)
Δ = 384
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:

$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8\sqrt{6}}{2*4}=\frac{-16-8\sqrt{6}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8\sqrt{6}}{2*4}=\frac{-16+8\sqrt{6}}{8}
$