4x^2+0.5x-0.01=0

Solution for 4x^2+0.5x-0.01=0 equation:

4x^2+0.5x-0.01=0

a = 4; b = 0.5; c = -0.01;
Δ = b2-4ac
Δ = 0.52-4·4·(-0.01)
Δ = 0.41
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{-(0.5)-\sqrt{0.41}}{2*4}=\frac{-0.5-\sqrt{0.41}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.5)+\sqrt{0.41}}{2*4}=\frac{-0.5+\sqrt{0.41}}{8}
$