5f2=2f+10

Solution for 5f2=2f+10 equation:

5f^2=2f+10

We move all terms to the left:

5f^2-(2f+10)=0

We get rid of parentheses

5f^2-2f-10=0

a = 5; b = -2; c = -10;
Δ = b2-4ac
Δ = -22-4·5·(-10)
Δ = 204
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:

$\sqrt{\Delta}=\sqrt{204}=\sqrt{4*51}=\sqrt{4}*\sqrt{51}=2\sqrt{51}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{51}}{2*5}=\frac{2-2\sqrt{51}}{10}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{51}}{2*5}=\frac{2+2\sqrt{51}}{10}
$