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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-4F^2+4F=0

a = -4; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-4)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*-4}=\frac{-8}{-8}
=1
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-4}=\frac{0}{-8}
=0
$