(2x+1)(4x+8)=10

Solution for (2x+1)(4x+8)=10 equation:

(2x+1)(4x+8)=10

We move all terms to the left:

(2x+1)(4x+8)-(10)=0

We multiply parentheses ..

(+8x^2+16x+4x+8)-10=0

We get rid of parentheses

8x^2+16x+4x+8-10=0

We add all the numbers together, and all the variables

8x^2+20x-2=0

a = 8; b = 20; c = -2;
Δ = b2-4ac
Δ = 202-4·8·(-2)
Δ = 464
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:

$\sqrt{\Delta}=\sqrt{464}=\sqrt{16*29}=\sqrt{16}*\sqrt{29}=4\sqrt{29}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{29}}{2*8}=\frac{-20-4\sqrt{29}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{29}}{2*8}=\frac{-20+4\sqrt{29}}{16}
$