F(x)=3x^2+x-10

Solution for F(x)=3x^2+x-10 equation:

(F)=3F^2+F-10

We move all terms to the left:

(F)-(3F^2+F-10)=0

We get rid of parentheses

-3F^2+F-F+10=0

We add all the numbers together, and all the variables

-3F^2+10=0

a = -3; b = 0; c = +10;
Δ = b2-4ac
Δ = 02-4·(-3)·10
Δ = 120
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:

$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{30}}{2*-3}=\frac{0-2\sqrt{30}}{-6}
=-\frac{2\sqrt{30}}{-6}
=-\frac{\sqrt{30}}{-3}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{30}}{2*-3}=\frac{0+2\sqrt{30}}{-6}
=\frac{2\sqrt{30}}{-6}
=\frac{\sqrt{30}}{-3}
$