2x2+0.3x-0.35=0

Solution for 2x2+0.3x-0.35=0 equation:

2x^2+0.3x-0.35=0

a = 2; b = 0.3; c = -0.35;
Δ = b2-4ac
Δ = 0.32-4·2·(-0.35)
Δ = 2.89
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{-(0.3)-\sqrt{2.89}}{2*2}=\frac{-0.3-\sqrt{2.89}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.3)+\sqrt{2.89}}{2*2}=\frac{-0.3+\sqrt{2.89}}{4}
$