3x^2=9+x

Solution for 3x^2=9+x equation:

3x^2=9+x

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

3x^2-x-9=0

We add all the numbers together, and all the variables

3x^2-1x-9=0

a = 3; b = -1; c = -9;
Δ = b2-4ac
Δ = -12-4·3·(-9)
Δ = 109
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{109}}{2*3}=\frac{1-\sqrt{109}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{109}}{2*3}=\frac{1+\sqrt{109}}{6}
$