2x^2+3x=1=0

Solution for 2x^2+3x=1=0 equation:

2x^2+3x=1=0

We move all terms to the left:

2x^2+3x-(1)=0

a = 2; b = 3; c = -1;
Δ = b2-4ac
Δ = 32-4·2·(-1)
Δ = 17
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{17}}{2*2}=\frac{-3-\sqrt{17}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{17}}{2*2}=\frac{-3+\sqrt{17}}{4}
$