F(x)=2x^2-16x+35

Solution for F(x)=2x^2-16x+35 equation:

(F)=2F^2-16F+35

We move all terms to the left:

(F)-(2F^2-16F+35)=0

We get rid of parentheses

-2F^2+F+16F-35=0

We add all the numbers together, and all the variables

-2F^2+17F-35=0

a = -2; b = 17; c = -35;
Δ = b2-4ac
Δ = 172-4·(-2)·(-35)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-3}{2*-2}=\frac{-20}{-4}
=+5
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+3}{2*-2}=\frac{-14}{-4}
=3+1/2
$