4x/8=2x^2

Solution for 4x/8=2x^2 equation:

4x/8=2x^2

We move all terms to the left:

4x/8-(2x^2)=0

determiningTheFunctionDomain
-2x^2+4x/8=0

We multiply all the terms by the denominator


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

Wy multiply elements

-16x^2+4x=0

a = -16; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-16)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*-16}=\frac{-8}{-32}
=1/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-16}=\frac{0}{-32}
=0
$