7(x^2-9)=x(x-3)

Solution for 7(x^2-9)=x(x-3) equation:

7(x^2-9)=x(x-3)

We move all terms to the left:

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

We multiply parentheses

7x^2-(x(x-3))-63=0


We calculate terms in parentheses: -(x(x-3)), so:

x(x-3)

We multiply parentheses

x^2-3x

Back to the equation:

-(x^2-3x)


We get rid of parentheses

7x^2-x^2+3x-63=0

We add all the numbers together, and all the variables

6x^2+3x-63=0

a = 6; b = 3; c = -63;
Δ = b2-4ac
Δ = 32-4·6·(-63)
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-39}{2*6}=\frac{-42}{12}
=-3+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+39}{2*6}=\frac{36}{12}
=3
$