F(x)=2x2-3/5

Solution for F(x)=2x2-3/5 equation:

(F)=2F^2-3/5

We move all terms to the left:

(F)-(2F^2-3/5)=0

We get rid of parentheses

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

We multiply all the terms by the denominator


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

Wy multiply elements

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

a = -10; b = 5; c = +3;
Δ = b2-4ac
Δ = 52-4·(-10)·3
Δ = 145
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{-(5)-\sqrt{145}}{2*-10}=\frac{-5-\sqrt{145}}{-20}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{145}}{2*-10}=\frac{-5+\sqrt{145}}{-20}
$