F(x)=2x^2-9x+10

Solution for F(x)=2x^2-9x+10 equation:

(F)=2F^2-9F+10

We move all terms to the left:

(F)-(2F^2-9F+10)=0

We get rid of parentheses

-2F^2+F+9F-10=0

We add all the numbers together, and all the variables

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

a = -2; b = 10; c = -10;
Δ = b2-4ac
Δ = 102-4·(-2)·(-10)
Δ = 20
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:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{5}}{2*-2}=\frac{-10-2\sqrt{5}}{-4}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{5}}{2*-2}=\frac{-10+2\sqrt{5}}{-4}
$