8x^+40x^2=O

Solution for 8x^+40x^2=O equation:

8x^+40x^2=

We move all terms to the left:

8x^+40x^2-()=0

We add all the numbers together, and all the variables

40x^2+8x=0

a = 40; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·40·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8}{2*40}=\frac{-16}{80}
=-1/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*40}=\frac{0}{80}
=0
$