-7f^2+11f+9=4-3f

Solution for -7f^2+11f+9=4-3f equation:

-7f^2+11f+9=4-3f

We move all terms to the left:

-7f^2+11f+9-(4-3f)=0

We add all the numbers together, and all the variables

-7f^2+11f-(-3f+4)+9=0

We get rid of parentheses

-7f^2+11f+3f-4+9=0

We add all the numbers together, and all the variables

-7f^2+14f+5=0

a = -7; b = 14; c = +5;
Δ = b2-4ac
Δ = 142-4·(-7)·5
Δ = 336
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:

$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-4\sqrt{21}}{2*-7}=\frac{-14-4\sqrt{21}}{-14}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+4\sqrt{21}}{2*-7}=\frac{-14+4\sqrt{21}}{-14}
$