50x^2+9x-0.9=0

Solution for 50x^2+9x-0.9=0 equation:

50x^2+9x-0.9=0

a = 50; b = 9; c = -0.9;
Δ = b2-4ac
Δ = 92-4·50·(-0.9)
Δ = 261
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{261}=\sqrt{9*29}=\sqrt{9}*\sqrt{29}=3\sqrt{29}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-3\sqrt{29}}{2*50}=\frac{-9-3\sqrt{29}}{100}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3\sqrt{29}}{2*50}=\frac{-9+3\sqrt{29}}{100}
$