2x^2-3x-9/8=0

Solution for 2x^2-3x-9/8=0 equation:

2x^2-3x-9/8=0

We multiply all the terms by the denominator


2x^2*8-3x*8-9=0

Wy multiply elements

16x^2-24x-9=0

a = 16; b = -24; c = -9;
Δ = b2-4ac
Δ = -242-4·16·(-9)
Δ = 1152
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:

$\sqrt{\Delta}=\sqrt{1152}=\sqrt{576*2}=\sqrt{576}*\sqrt{2}=24\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-24\sqrt{2}}{2*16}=\frac{24-24\sqrt{2}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+24\sqrt{2}}{2*16}=\frac{24+24\sqrt{2}}{32}
$