(2(2x+5)-11)(2x+5)=40

Solution for (2(2x+5)-11)(2x+5)=40 equation:

(2(2x+5)-11)(2x+5)=40

We move all terms to the left:

(2(2x+5)-11)(2x+5)-(40)=0


We calculate terms in parentheses: +(2(2x+5)-11)(2x+5), so:

2(2x+5)-11)(2x+5

We multiply parentheses

4x-11)(2x+10+5

We add all the numbers together, and all the variables

4x-11)(2x+15

Back to the equation:

+(4x-11)(2x+15)


We multiply parentheses ..

(+8x^2+60x-22x-165)-40=0

We get rid of parentheses

8x^2+60x-22x-165-40=0

We add all the numbers together, and all the variables

8x^2+38x-205=0

a = 8; b = 38; c = -205;
Δ = b2-4ac
Δ = 382-4·8·(-205)
Δ = 8004
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:

$\sqrt{\Delta}=\sqrt{8004}=\sqrt{4*2001}=\sqrt{4}*\sqrt{2001}=2\sqrt{2001}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(38)-2\sqrt{2001}}{2*8}=\frac{-38-2\sqrt{2001}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(38)+2\sqrt{2001}}{2*8}=\frac{-38+2\sqrt{2001}}{16}
$