2y2+3.5y=1

Solution for 2y2+3.5y=1 equation:

2y^2+3.5y=1

We move all terms to the left:

2y^2+3.5y-(1)=0

a = 2; b = 3.5; c = -1;
Δ = b2-4ac
Δ = 3.52-4·2·(-1)
Δ = 20.25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3.5)-\sqrt{20.25}}{2*2}=\frac{-3.5-\sqrt{20.25}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3.5)+\sqrt{20.25}}{2*2}=\frac{-3.5+\sqrt{20.25}}{4}
$