Question

the quadratic formula
by starting with $ax^{2}+bx + c = 0$ (which can stand for any quadratic equation) and completing the square, we can get formulas for the solutions.
$x=\frac{-b+sqrt{b^{2}-4ac}}{2a}$
$x=\frac{-b-sqrt{b^{2}-4ac}}{2a}$
the two formulas are exactly the same except for the sign in the middle. we usually combine them into one by writing $pm$ (\plus or minus\) in front of the square root sign.
$x=\frac{-bpmsqrt{b^{2}-4ac}}{2a}$
this is called the quadratic formula. lets use it to find the solutions for the last equation on page 25.
first we rewrite the equation in standard form ($ax^{2}+bx + c = 0$).
then we substitute the values of $a$, $b$ and $c$ in the formula.
then we simplify the result.
$2x^{2}-12x + 14 = 0$
$\textcircled{2}x^{2}+\textcircled{-12}x+\textcircled{14}=0$
$x=\frac{-(-12)pmsqrt{(-12)^{2}-4(2)(14)}}{2(2)}$
$x=\frac{12pmsqrt{144 - 112}}{4}$
$x=\frac{12pmsqrt{32}}{4}$
$x=\frac{12pm4sqrt{2}}{4}$
$x=\frac{12}{4}pm\frac{4sqrt{2}}{4}$
$x = 3pmsqrt{2}$

Explanation:

Step1: Identify coefficients

For the equation $2x^{2}-12x + 14=0$, $a = 2$, $b=-12$, $c = 14$.

Step2: Substitute into quadratic formula

$x=\frac{-(-12)\pm\sqrt{(-12)^{2}-4(2)(14)}}{2(2)}=\frac{12\pm\sqrt{144 - 112}}{4}$

Step3: Simplify square - root part

$\sqrt{144 - 112}=\sqrt{32}=4\sqrt{2}$, so $x=\frac{12\pm4\sqrt{2}}{4}$

Step4: Simplify the fraction

$x=\frac{12}{4}\pm\frac{4\sqrt{2}}{4}=3\pm\sqrt{2}$

Answer:

$x = 3\pm\sqrt{2}$