SNSWER
. x = 2
F. x = -1
EXPLANATION
e can rewrite this equation as a quadratic frunction equal to zero.
First apply the binomial squared on the left side:
[tex](x-2)^2=x^2-4x+4[/tex]
This is to write it in standard form. Now we have the equation:
[tex]x^2-4x+4=-3x+6[/tex]
We can add 3x on both sides of the equation:
[tex]\begin{gathered} x^2-4x+3x+4=-3x+3x+6 \\ x^2-x+4=6 \end{gathered}[/tex]
And subtract 6 from both sides:
[tex]\begin{gathered} x^2-x+4-6=6-6 \\ x^2-x-2=0 \end{gathered}[/tex]
Now we can use the quadratic formula to solve this for x:
[tex]\begin{gathered} ax^2+bx+c=0 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}[/tex]
In our equation a = 1, b = -1 and c = -2:
[tex]\begin{gathered} x=\frac{1\pm\sqrt[]{1^2+4\cdot2\cdot1}}{2\cdot1} \\ x=\frac{1\pm\sqrt[]{1+8}}{2} \\ x=\frac{1\pm\sqrt[]{9}}{2} \\ x=\frac{1\pm3}{2} \\ x_1=\frac{1+3}{2}=\frac{4}{2}=2_{} \\ x_2=\frac{1-3}{2}=\frac{-2}{2}=-1 \end{gathered}[/tex]
Therefore the solutions to the given equation are x = 2 and x = -1
