.01x^2+.7x+6.3=0

Solution for .01x^2+.7x+6.3=0 equation:

.01x^2+.7x+6.3=0

a = .01; b = .7; c = +6.3;
Δ = b2-4ac
Δ = .72-4·.01·6.3
Δ = 0.238
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{-(.7)-\sqrt{0.238}}{2*.01}=\frac{-0.7-\sqrt{0.238}}{0.02}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(.7)+\sqrt{0.238}}{2*.01}=\frac{-0.7+\sqrt{0.238}}{0.02}
$