(9x-5)(x+1)=0

Solution for (9x-5)(x+1)=0 equation:

(9x-5)(x+1)=0

We multiply parentheses ..

(+9x^2+9x-5x-5)=0

We get rid of parentheses

9x^2+9x-5x-5=0

We add all the numbers together, and all the variables

9x^2+4x-5=0

a = 9; b = 4; c = -5;
Δ = b2-4ac
Δ = 42-4·9·(-5)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-14}{2*9}=\frac{-18}{18}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+14}{2*9}=\frac{10}{18}
=5/9
$