(-2f2-3f-5)+(-2f2-3f+8)=0

Solution for (-2f2-3f-5)+(-2f2-3f+8)=0 equation:

(-2f^2-3f-5)+(-2f^2-3f+8)=0

We get rid of parentheses

-2f^2-2f^2-3f-3f-5+8=0

We add all the numbers together, and all the variables

-4f^2-6f+3=0

a = -4; b = -6; c = +3;
Δ = b2-4ac
Δ = -62-4·(-4)·3
Δ = 84
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{84}=\sqrt{4*21}=\sqrt{4}*\sqrt{21}=2\sqrt{21}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{21}}{2*-4}=\frac{6-2\sqrt{21}}{-8}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{21}}{2*-4}=\frac{6+2\sqrt{21}}{-8}
$