Solve using the quadratic formula. $\newline$ $6f^2 + 6f - 6 = 0$ $\newline$ Write your answers as integers, proper or improper fractions in simplest form, or decimals rounded to the nearest hundredth. $\newline$ $f =$ _ or $f =$ _

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Algebra 1

Solve a quadratic equation using the quadratic formula

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Q. Solve using the quadratic formula.

\newline

6f^2 + 6f - 6 = 0

\newline

Write your answers as integers, proper or improper fractions in simplest form, or decimals rounded to the nearest hundredth.

\newline

f =

_____ or

f =

_____

Identify values of quadratic equation: Identify the values of $a$ , $b$ , and $c$ in the quadratic equation $6f^2 + 6f - 6 = 0$ . The quadratic equation is in the form $af^2 + bf + c = 0$ , so by comparison: $a = 6$ $b = 6$ $c = -6$
Substitute values into formula: Substitute the values of $a$ , $b$ , and $c$ into the quadratic formula $f = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ . The quadratic formula is $f = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , so we substitute: $f = \frac{-(6) \pm \sqrt{(6)^2 - 4\cdot6\cdot(-6)}}{2\cdot6}$
Simplify discriminant: Simplify the expression under the square root (the discriminant). $\newline$ Calculate the discriminant: $(6)^2 - 4\cdot6\cdot(-6)$ $\newline$ = $36 + 144$ $\newline$ = $180$
Continue simplifying formula: Continue simplifying the quadratic formula with the calculated discriminant. $f = \frac{-6 \pm \sqrt{180}}{12}$
Simplify square root: Simplify the square root of the discriminant, if possible. $\sqrt{180}$ can be simplified because $180 = 36 \times 5$ , and $36$ is a perfect square. $\sqrt{180} = \sqrt{36 \times 5} = \sqrt{36} \times \sqrt{5} = 6 \times \sqrt{5}$
Substitute simplified root: Substitute the simplified square root back into the quadratic formula. $\newline$ $f = \frac{{-6 \pm 6 \times \sqrt{5}}}{{12}}$
Factor out common terms: Simplify the quadratic formula by factoring out common terms. Both terms in the numerator have a common factor of $6$ . $f = \frac{6(-1 \pm \sqrt{5})}{6\times2} f = \frac{-1 \pm \sqrt{5}}{2}$
Identify possible values for f: Identify the two possible values for f. $\newline$ $f = \frac{-1 + \sqrt{5}}{2}$ or $f = \frac{-1 - \sqrt{5}}{2}$
Round values to nearest hundredth: If necessary, round the values of $f$ to the nearest hundredth.f \approx (\-1 + 2.24) / 2 or f \approx (\-1 - 2.24) / 2 $f \approx 1.24 / 2$ or f \approx \-3.24 / 2 $f \approx 0.62$ or f \approx \-1.62