F(x)=3x^2+30x-65

Solution for F(x)=3x^2+30x-65 equation:

(F)=3F^2+30F-65

We move all terms to the left:

(F)-(3F^2+30F-65)=0

We get rid of parentheses

-3F^2+F-30F+65=0

We add all the numbers together, and all the variables

-3F^2-29F+65=0

a = -3; b = -29; c = +65;
Δ = b2-4ac
Δ = -292-4·(-3)·65
Δ = 1621
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-\sqrt{1621}}{2*-3}=\frac{29-\sqrt{1621}}{-6}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+\sqrt{1621}}{2*-3}=\frac{29+\sqrt{1621}}{-6}
$