2x^2-x=17/4

Solution for 2x^2-x=17/4 equation:

2x^2-x=17/4

We move all terms to the left:

2x^2-x-(17/4)=0

We add all the numbers together, and all the variables

2x^2-x-(+17/4)=0

We add all the numbers together, and all the variables

2x^2-1x-(+17/4)=0

We get rid of parentheses

2x^2-1x-17/4=0

We multiply all the terms by the denominator


2x^2*4-1x*4-17=0

Wy multiply elements

8x^2-4x-17=0

a = 8; b = -4; c = -17;
Δ = b2-4ac
Δ = -42-4·8·(-17)
Δ = 560
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:

$\sqrt{\Delta}=\sqrt{560}=\sqrt{16*35}=\sqrt{16}*\sqrt{35}=4\sqrt{35}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{35}}{2*8}=\frac{4-4\sqrt{35}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{35}}{2*8}=\frac{4+4\sqrt{35}}{16}
$