f2+25f+150=0

Solution for f2+25f+150=0 equation:

f2+25f+150=0

We add all the numbers together, and all the variables

f^2+25f+150=0

a = 1; b = 25; c = +150;
Δ = b2-4ac
Δ = 252-4·1·150
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-5}{2*1}=\frac{-30}{2}
=-15
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+5}{2*1}=\frac{-20}{2}
=-10
$