2x^2-3x-25/8=0

Solution for 2x^2-3x-25/8=0 equation:

2x^2-3x-25/8=0

We multiply all the terms by the denominator


2x^2*8-3x*8-25=0

Wy multiply elements

16x^2-24x-25=0

a = 16; b = -24; c = -25;
Δ = b2-4ac
Δ = -242-4·16·(-25)
Δ = 2176
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:

$\sqrt{\Delta}=\sqrt{2176}=\sqrt{64*34}=\sqrt{64}*\sqrt{34}=8\sqrt{34}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-8\sqrt{34}}{2*16}=\frac{24-8\sqrt{34}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+8\sqrt{34}}{2*16}=\frac{24+8\sqrt{34}}{32}
$