F(x)=2x^2-4x^2-5x+3

Solution for F(x)=2x^2-4x^2-5x+3 equation:

(F)=2F^2-4F^2-5F+3

We move all terms to the left:

(F)-(2F^2-4F^2-5F+3)=0

We get rid of parentheses

-2F^2+4F^2+5F+F-3=0

We add all the numbers together, and all the variables

2F^2+6F-3=0

a = 2; b = 6; c = -3;
Δ = b2-4ac
Δ = 62-4·2·(-3)
Δ = 60
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:

$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{15}}{2*2}=\frac{-6-2\sqrt{15}}{4}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{15}}{2*2}=\frac{-6+2\sqrt{15}}{4}
$