(8f-9)(f+5)=0

Solution for (8f-9)(f+5)=0 equation:

(8f-9)(f+5)=0

We multiply parentheses ..

(+8f^2+40f-9f-45)=0

We get rid of parentheses

8f^2+40f-9f-45=0

We add all the numbers together, and all the variables

8f^2+31f-45=0

a = 8; b = 31; c = -45;
Δ = b2-4ac
Δ = 312-4·8·(-45)
Δ = 2401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2401}=49$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-49}{2*8}=\frac{-80}{16}
=-5
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+49}{2*8}=\frac{18}{16}
=1+1/8
$