(12f^2-9f=15)-(-4f^2-3f-6)

Solution for (12f^2-9f=15)-(-4f^2-3f-6) equation:

(12f^2-9f=15)-(-4f^2-3f-6)

We move all terms to the left:

(12f^2-9f-(15)-(-4f^2-3f-6))=0


We calculate terms in parentheses: +(12f^2-9f-15-(-4f^2-3f-6)), so:

12f^2-9f-15-(-4f^2-3f-6)

determiningTheFunctionDomain
12f^2-(-4f^2-3f-6)-9f-15

We get rid of parentheses

12f^2+4f^2+3f-9f+6-15

We add all the numbers together, and all the variables

16f^2-6f-9

Back to the equation:

+(16f^2-6f-9)


We get rid of parentheses

16f^2-6f-9=0

a = 16; b = -6; c = -9;
Δ = b2-4ac
Δ = -62-4·16·(-9)
Δ = 612
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:

$\sqrt{\Delta}=\sqrt{612}=\sqrt{36*17}=\sqrt{36}*\sqrt{17}=6\sqrt{17}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6\sqrt{17}}{2*16}=\frac{6-6\sqrt{17}}{32}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6\sqrt{17}}{2*16}=\frac{6+6\sqrt{17}}{32}
$