(3+x)-(2x2)+6=9

Solution for (3+x)-(2x2)+6=9 equation:

(3+x)-(2x^2)+6=9

We move all terms to the left:

(3+x)-(2x^2)+6-(9)=0

determiningTheFunctionDomain
-2x^2+(3+x)+6-9=0

We add all the numbers together, and all the variables

-2x^2+(x+3)+6-9=0

We add all the numbers together, and all the variables

-2x^2+(x+3)-3=0

We get rid of parentheses

-2x^2+x+3-3=0

We add all the numbers together, and all the variables

-2x^2+x=0

a = -2; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-2)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-2}=\frac{-2}{-4}
=1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-2}=\frac{0}{-4}
=0
$