2y^2+5y-3/4=0

Solution for 2y^2+5y-3/4=0 equation:

2y^2+5y-3/4=0

We multiply all the terms by the denominator


2y^2*4+5y*4-3=0

Wy multiply elements

8y^2+20y-3=0

a = 8; b = 20; c = -3;
Δ = b2-4ac
Δ = 202-4·8·(-3)
Δ = 496
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{496}=\sqrt{16*31}=\sqrt{16}*\sqrt{31}=4\sqrt{31}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{31}}{2*8}=\frac{-20-4\sqrt{31}}{16}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{31}}{2*8}=\frac{-20+4\sqrt{31}}{16}
$