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

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

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

We move all terms to the left:

2y^2+5y-4y-(3)=0

We add all the numbers together, and all the variables

2y^2+y-3=0

a = 2; b = 1; c = -3;
Δ = b2-4ac
Δ = 12-4·2·(-3)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-5}{2*2}=\frac{-6}{4}
=-1+1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+5}{2*2}=\frac{4}{4}
=1
$