Respuesta :
We are given the following quadratic equation
[tex]4x^2-30x=12+10x[/tex]Let us solve this equation by completing the square method.
First of all, simplify the equation a bit
[tex]\begin{gathered} 4x^2-30x=12+10x \\ 4x^2-30x-10x-12=0 \\ 4x^2-40x-12=0 \end{gathered}[/tex]Step 1:
Divide the equation by 4 (to make the coefficient of the square term 1)
[tex]\begin{gathered} \frac{4x^2-40x-12}{4}=\frac{0}{4} \\ x^2-10x-3=0 \end{gathered}[/tex]Step 2:
Move the constant term to the right side
[tex]x^2-10x=3[/tex]Step 3:
Add half of the square of the coefficient of x-term to both sides of the equation
The coefficient of x-term is -10
[tex]x^2-10x+(\frac{-10}{2})^2=3+(\frac{-10}{2})^2[/tex]Simplify
[tex]\begin{gathered} x^2-10x+(-5)^2=3+(-5)^2 \\ x^2-10x+25^{}=3+25^{} \\ x^2-10x+25^{}=28 \end{gathered}[/tex]Step 4:
Apply the difference of squares formula on the left side of the equation
[tex](a-b)^2=a^2-2ab+b^2[/tex]In this case, a = x and b = 5
[tex](x-5)^2=28[/tex]Step 5:
Now we can solve the equation for x.
Take square root on both sides of the equation
[tex]\begin{gathered} \sqrt[]{\mleft(x-5\mright)^2}=\sqrt[]{28} \\ (x-5)=\pm\sqrt[]{28} \\ (x-5)=\pm2\sqrt[]{7} \end{gathered}[/tex]Step 6:
Find the two possible solutions of x
[tex]\begin{gathered} x-5=2\sqrt[]{7}\quad and\quad x-5=-2\sqrt[]{7} \\ x=5+2\sqrt[]{7}\quad and\quad x=5-2\sqrt[]{7} \\ x=5+5.29\quad and\quad x=5-5.29 \\ x=10.29\quad and\quad x=-0.29 \end{gathered}[/tex]Therefore, the solution of the given quadratic equation is
[tex]\begin{gathered} x=(5+2\sqrt[]{7},5-2\sqrt[]{7}) \\ x=(10.29,-0.29) \end{gathered}[/tex]