9x^2+0.8x-0.12=0

Solution for 9x^2+0.8x-0.12=0 equation:

9x^2+0.8x-0.12=0

a = 9; b = 0.8; c = -0.12;
Δ = b2-4ac
Δ = 0.82-4·9·(-0.12)
Δ = 4.96
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{-(0.8)-\sqrt{4.96}}{2*9}=\frac{-0.8-\sqrt{4.96}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.8)+\sqrt{4.96}}{2*9}=\frac{-0.8+\sqrt{4.96}}{18}
$