F(x)=3x^2+5x-10

Solution for F(x)=3x^2+5x-10 equation:

(F)=3F^2+5F-10

We move all terms to the left:

(F)-(3F^2+5F-10)=0

We get rid of parentheses

-3F^2+F-5F+10=0

We add all the numbers together, and all the variables

-3F^2-4F+10=0

a = -3; b = -4; c = +10;
Δ = b2-4ac
Δ = -42-4·(-3)·10
Δ = 136
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:

$\sqrt{\Delta}=\sqrt{136}=\sqrt{4*34}=\sqrt{4}*\sqrt{34}=2\sqrt{34}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{34}}{2*-3}=\frac{4-2\sqrt{34}}{-6}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{34}}{2*-3}=\frac{4+2\sqrt{34}}{-6}
$