8x(2x+5)=16x+40

Solution for 8x(2x+5)=16x+40 equation:

8x(2x+5)=16x+40

We move all terms to the left:

8x(2x+5)-(16x+40)=0

We multiply parentheses

16x^2+40x-(16x+40)=0

We get rid of parentheses

16x^2+40x-16x-40=0

We add all the numbers together, and all the variables

16x^2+24x-40=0

a = 16; b = 24; c = -40;
Δ = b2-4ac
Δ = 242-4·16·(-40)
Δ = 3136
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3136}=56$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-56}{2*16}=\frac{-80}{32}
=-2+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+56}{2*16}=\frac{32}{32}
=1
$