(3/10)x^2-(2/5)x+(7/10)=0

Solution for (3/10)x^2-(2/5)x+(7/10)=0 equation:

(3/10)x^2-(2/5)x+(7/10)=0


Domain of the equation: 10)x^2!=0

x!=0/1

x!=0

x∈R



Domain of the equation: 5)x!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+3/10)x^2-(+2/5)x+(+7/10)=0

We multiply parentheses

-2x^2+(+3/10)x^2+(+7/10)=0

We get rid of parentheses

-2x^2+(+3/10)x^2+7/10=0

We calculate fractions


-2x^2=0

a = -2; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·(-2)·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{0}{-4}=0$