F(x)=10x^2-140x+8

Solution for F(x)=10x^2-140x+8 equation:

(F)=10F^2-140F+8

We move all terms to the left:

(F)-(10F^2-140F+8)=0

We get rid of parentheses

-10F^2+F+140F-8=0

We add all the numbers together, and all the variables

-10F^2+141F-8=0

a = -10; b = 141; c = -8;
Δ = b2-4ac
Δ = 1412-4·(-10)·(-8)
Δ = 19561
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(141)-\sqrt{19561}}{2*-10}=\frac{-141-\sqrt{19561}}{-20}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(141)+\sqrt{19561}}{2*-10}=\frac{-141+\sqrt{19561}}{-20}
$