F(m)=4m^2-48m+80

Solution for F(m)=4m^2-48m+80 equation:

(F)=4F^2-48F+80

We move all terms to the left:

(F)-(4F^2-48F+80)=0

We get rid of parentheses

-4F^2+F+48F-80=0

We add all the numbers together, and all the variables

-4F^2+49F-80=0

a = -4; b = 49; c = -80;
Δ = b2-4ac
Δ = 492-4·(-4)·(-80)
Δ = 1121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(49)-\sqrt{1121}}{2*-4}=\frac{-49-\sqrt{1121}}{-8}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(49)+\sqrt{1121}}{2*-4}=\frac{-49+\sqrt{1121}}{-8}
$