(4x+9)(x-5)=0

Solution for (4x+9)(x-5)=0 equation:

(4x+9)(x-5)=0

We multiply parentheses ..

(+4x^2-20x+9x-45)=0

We get rid of parentheses

4x^2-20x+9x-45=0

We add all the numbers together, and all the variables

4x^2-11x-45=0

a = 4; b = -11; c = -45;
Δ = b2-4ac
Δ = -112-4·4·(-45)
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-29}{2*4}=\frac{-18}{8}
=-2+1/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+29}{2*4}=\frac{40}{8}
=5
$