2x-3^2-4x^2/2x+8x/4=-9

Solution for 2x-3^2-4x^2/2x+8x/4=-9 equation:

2x-3^2-4x^2/2x+8x/4=-9

We move all terms to the left:

2x-3^2-4x^2/2x+8x/4-(-9)=0


Domain of the equation: 2x!=0

x!=0/2

x!=0

x∈R


We add all the numbers together, and all the variables

-4x^2/2x+2x+8x/4=0

We calculate fractions


(-16x^2)/8x+16x^2/8x+2x=0

We multiply all the terms by the denominator


(-16x^2)+16x^2+2x*8x=0

We add all the numbers together, and all the variables

16x^2+(-16x^2)+2x*8x=0

Wy multiply elements

16x^2+(-16x^2)+16x^2=0

We get rid of parentheses

16x^2-16x^2+16x^2=0

We add all the numbers together, and all the variables

16x^2=0

a = 16; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·16·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{0}{32}=0$