suppose you are a solving a quadratic equation and this is your work so far...\n$x^{2}-12x…

suppose you are a solving a quadratic equation and this is your work so far...\n$x^{2}-12x - 28=0$\n$x=\frac{12pmsqrt{(-12)^{2}-4(1)(-28)}}{2(1)}$\n$x=\frac{12pmsqrt{144 + 112}}{2}$\n$x=\frac{12pmsqrt{256}}{2}$\n...keep going! what will the solutions to this quadratic equation be?\n$x=-1$ and $x = 13$\n$x=-2$ and $x = 14$\n$x=133.5$ and $x=-122.5$\n$x=2$ and $x=-14$

suppose you are a solving a quadratic equation and this is your work so far...\n$x^{2}-12x - 28=0$\n$x=\frac{12pmsqrt{(-12)^{2}-4(1)(-28)}}{2(1)}$\n$x=\frac{12pmsqrt{144 + 112}}{2}$\n$x=\frac{12pmsqrt{256}}{2}$\n...keep going! what will the solutions to this quadratic equation be?\n$x=-1$ and $x = 13$\n$x=-2$ and $x = 14$\n$x=133.5$ and $x=-122.5$\n$x=2$ and $x=-14$

Answer

Explanation:

Step1: Evaluate square - root

Since $\sqrt{256}=16$, the equation $x = \frac{12\pm\sqrt{256}}{2}$ becomes $x=\frac{12\pm16}{2}$.

Step2: Calculate two solutions

For the plus - sign: $x=\frac{12 + 16}{2}=\frac{28}{2}=14$. For the minus - sign: $x=\frac{12-16}{2}=\frac{-4}{2}=-2$.

Answer:

x = - 2 and x = 14