Question

Hey! If you can help here it is!

Answer 1

Answer:

$x=3,x=-1$

Step-by-step explanation:

Start with:

$x(x-1)-3(x-3)+3(x-2)=x+6$

Distribute the values on the outside of each of the parentheses.

$x^2-x-3x+9+3x-6=x+6$

Combine like terms.

$x^2-x+3=x+6$

Subtract $6$ from both sides of the equation.

$x^2-x-3=x$

Subtract $x$ from both sides of the equation.

$x^2-2x-3=0$

Now, we need to use our quadratic equation formula:

$ax^2+bx+c=0$

$x= \frac{-b+/-\sqrt{b^2-4ac}}{2a}$

Identify your values.

$a=1\\b=-2\\c=-3$

Substitute.

(Solve positive)

$x= \frac{-(-2)+\sqrt{(-2)^2-4(1)(-3)}}{2(1)}$

Solve.

$x= \frac{2+\sqrt{(16}}{2}$

Find the square root of $16$.

$x=\frac{2+4}{2}$

Add.

$x=\frac{6}{2}$

Simplify.

$x=3$

~

(Solve negative)

$x= \frac{-(-2)-\sqrt{(-2)^2-4(1)(-3)}}{2(1)}$

Solve.

$x= \frac{2-\sqrt{(16}}{2}$

Find the square root of $16$.

$x=\frac{2-4}{2}$

Subtract.

$x=\frac{-2}{2}$

Simplify.

$x=-1$