0.01x^2+0.7x+6.2=0

Solution for 0.01x^2+0.7x+6.2=0 equation:

0.01x^2+0.7x+6.2=0

a = 0.01; b = 0.7; c = +6.2;
Δ = b2-4ac
Δ = 0.72-4·0.01·6.2
Δ = 0.242
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{-(0.7)-\sqrt{0.242}}{2*0.01}=\frac{-0.7-\sqrt{0.242}}{0.02}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.7)+\sqrt{0.242}}{2*0.01}=\frac{-0.7+\sqrt{0.242}}{0.02}
$