3x^2-35=x(x+9)

Solution for 3x^2-35=x(x+9) equation:

3x^2-35=x(x+9)

We move all terms to the left:

3x^2-35-(x(x+9))=0


We calculate terms in parentheses: -(x(x+9)), so:

x(x+9)

We multiply parentheses

x^2+9x

Back to the equation:

-(x^2+9x)


We get rid of parentheses

3x^2-x^2-9x-35=0

We add all the numbers together, and all the variables

2x^2-9x-35=0

a = 2; b = -9; c = -35;
Δ = b2-4ac
Δ = -92-4·2·(-35)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-19}{2*2}=\frac{-10}{4}
=-2+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+19}{2*2}=\frac{28}{4}
=7
$