2x2+0.1x-0.03=0

Solution for 2x2+0.1x-0.03=0 equation:

2x^2+0.1x-0.03=0

a = 2; b = 0.1; c = -0.03;
Δ = b2-4ac
Δ = 0.12-4·2·(-0.03)
Δ = 0.25
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{-(0.1)-\sqrt{0.25}}{2*2}=\frac{-0.1-\sqrt{0.25}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.1)+\sqrt{0.25}}{2*2}=\frac{-0.1+\sqrt{0.25}}{4}
$