(2x*x)+x-(1/8)=0

Solution for (2x*x)+x-(1/8)=0 equation:

(2x*x)+x-(1/8)=0

We add all the numbers together, and all the variables

(+2x*x)+x-(+1/8)=0

We add all the numbers together, and all the variables

x+(+2x*x)-(+1/8)=0

We get rid of parentheses

x+2x*x-1/8=0

We multiply all the terms by the denominator


x*8+(2x*x)*8-1=0

We add all the numbers together, and all the variables

x*8+(+2x*x)*8-1=0

We multiply parentheses

16x^2+x*8-1=0

Wy multiply elements

16x^2+8x-1=0

a = 16; b = 8; c = -1;
Δ = b2-4ac
Δ = 82-4·16·(-1)
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{2}}{2*16}=\frac{-8-8\sqrt{2}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{2}}{2*16}=\frac{-8+8\sqrt{2}}{32}
$