0=2x^2-25/8-x

Solution for 0=2x^2-25/8-x equation:

0=2x^2-25/8-x

We move all terms to the left:

0-(2x^2-25/8-x)=0


Domain of the equation: 8-x)!=0

We move all terms containing x to the left, all other terms to the right

-x)!=-8

x!=-8/1

x!=-8

x∈R


We add all the numbers together, and all the variables

-(2x^2-25/8-x)=0

We get rid of parentheses

-2x^2+x+25/8=0

We multiply all the terms by the denominator


-2x^2*8+x*8+25=0

Wy multiply elements

-16x^2+8x+25=0

a = -16; b = 8; c = +25;
Δ = b2-4ac
Δ = 82-4·(-16)·25
Δ = 1664
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:

$\sqrt{\Delta}=\sqrt{1664}=\sqrt{64*26}=\sqrt{64}*\sqrt{26}=8\sqrt{26}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{26}}{2*-16}=\frac{-8-8\sqrt{26}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{26}}{2*-16}=\frac{-8+8\sqrt{26}}{-32}
$