Given the equation:
![x^2-4x+12=6x-2](https://img.qammunity.org/2023/formulas/mathematics/college/z4wlnbur48eu7a9fb1sw8tsc76af3e5o9x.png)
Let's find the solution to the given equation.
Move all terms to the left and equate to zero.
Ad 2 and subtract 6x to both sides:
![\begin{gathered} x^2-4x-6x+12+2=6x-6x-2+2_{} \\ \\ x^2-10x+14=0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/5bjsi0jot6naqqifkitobuxibnem1obiov.png)
Let's solve usingbthe quadratic formula:
![x=\frac{-b^{}\pm\sqrt[]{b^2-4ax}}{2a}](https://img.qammunity.org/2023/formulas/mathematics/college/iom1qqm92z5orc4b8064r4as6kmsoiud7k.png)
Apply the general formula to find the values of a, b, and c:
![\begin{gathered} ax^2+bx+c=0 \\ \\ x^2-10x+14=0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/3wumsnxeoybge27x0hhupqsv5xwzyr28ga.png)
Thus, we have:
a = 1
b = -10
c = 14
Substitute the values into the quadratic formula:
![\begin{gathered} x=\frac{-(-10)\pm\sqrt[]{-10^2+4(1)(14)}}{2(1)} \\ \\ x=\frac{10\pm\sqrt[]{100+56_{}}}{2} \\ \\ x=\frac{10\pm\sqrt[]{44}}{2} \\ \\ x=\frac{10\pm\sqrt[]{4\ast11}}{2} \\ \\ x=\frac{10\pm\sqrt[]{2^2\ast11}}{2} \\ \\ x=\frac{10\pm2\sqrt[]{11}}{2} \\ \\ x=5\pm\sqrt[]{11} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/lgu879u9iqjaeaewajio4vs2lg0qdh43d9.png)
Solving further:
![\begin{gathered} x=5+\sqrt[]{11},\text{ 5-}\sqrt[]{11} \\ \\ x=5+3.317,\text{ 5-3.31}7 \\ \\ x=8.317,\text{ 1.683} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/oxywa8rnhda2xp5x736h65fs1pcgyo1vr8.png)
Therefore, the solutions are:
x = 8.317, 1.683
ANSWER:
8.137 and 1.683