Solve Using the Quadratic Formula x^2-14x+48=0

Problem

$x^2-14x+48=0$

Solution

1. Identify the coefficients $a$ $b$ and $c$ from the quadratic equation in the form $a{x}^2+bx+c=0$

$a=1$

$b=-14$

$c=48$

2. State the quadratic formula used to find the roots of the equation.

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

3. Substitute the identified values into the quadratic formula.

$x=\frac{-(-14)\pm \sqrt{{(-14)}^2-4\left(1\right)\left(48\right)}}{2\left(1\right)}$

4. Simplify the expression inside the square root (the discriminant).

$x=\frac{14\pm \sqrt{196-192}}{2}$

$x=\frac{14\pm \sqrt{4}}{2}$

5. Evaluate the square root and simplify the fraction to find the two possible values for $x$

$x=\frac{14\pm 2}{2}$

6. Calculate the first solution using the plus sign.

$x=\frac{14+2}{2}=\frac{16}{2}=8$

7. Calculate the second solution using the minus sign.

$x=\frac{14-2}{2}=\frac{12}{2}=6$

Final Answer

$x=6,8$

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