F(x)=-2x^2+3x+38/4

Solution for F(x)=-2x^2+3x+38/4 equation:

(F)=-2F^2+3F+38/4

We move all terms to the left:

(F)-(-2F^2+3F+38/4)=0

We get rid of parentheses

2F^2-3F+F-38/4=0

We multiply all the terms by the denominator


2F^2*4-3F*4+F*4-38=0

Wy multiply elements

8F^2-12F+4F-38=0

We add all the numbers together, and all the variables

8F^2-8F-38=0

a = 8; b = -8; c = -38;
Δ = b2-4ac
Δ = -82-4·8·(-38)
Δ = 1280
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:

$\sqrt{\Delta}=\sqrt{1280}=\sqrt{256*5}=\sqrt{256}*\sqrt{5}=16\sqrt{5}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-16\sqrt{5}}{2*8}=\frac{8-16\sqrt{5}}{16}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+16\sqrt{5}}{2*8}=\frac{8+16\sqrt{5}}{16}
$