-10+-13f+3f^2=0

Solution for -10+-13f+3f^2=0 equation:

-10+-13f+3f^2=0

We add all the numbers together, and all the variables

3f^2-13f=0

a = 3; b = -13; c = 0;
Δ = b2-4ac
Δ = -132-4·3·0
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-13}{2*3}=\frac{0}{6}
=0
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+13}{2*3}=\frac{26}{6}
=4+1/3
$