10f+15=10f2

Solution for 10f+15=10f2 equation:

10f+15=10f^2

We move all terms to the left:

10f+15-(10f^2)=0

determiningTheFunctionDomain
-10f^2+10f+15=0

a = -10; b = 10; c = +15;
Δ = b2-4ac
Δ = 102-4·(-10)·15
Δ = 700
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:

$\sqrt{\Delta}=\sqrt{700}=\sqrt{100*7}=\sqrt{100}*\sqrt{7}=10\sqrt{7}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10\sqrt{7}}{2*-10}=\frac{-10-10\sqrt{7}}{-20}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10\sqrt{7}}{2*-10}=\frac{-10+10\sqrt{7}}{-20}
$