Mathematics, 07.07.2019 02:30 starreee

Which shows the correct substitution of the values a, b, and c from the equation 0 = – 3x2 – 2x + 6 into the quadratic formula? quadratic formula: x =

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$x_{1,2}=\frac{-(-2)\pm\sqrt{(-2)^2-4\cdot (-3)\cdot 6} }{2(-3)}$ shows correct substitution of the values a, b, and c from the given quadratic equation -3x^2-2x+6=0 into quadratic formula.

Step-by-step explanation:

Given: The quadratic equation -3x^2-2x+6=0

We have to show the correct substitution of the values a, b, and c from the given quadratic equation -3x^2-2x+6=0 into quadratic formula.

The standard form of quadratic equation is ax^2+bx+c=0 then the solution of quadratic equation using quadratic formula is given as $x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac} }{2a}$

Consider the given quadratic equation -3x^2-2x+6=0

Comparing with general quadratic equation, we have

a = -3 , b = -2 , c = 6

Substitute in quadratic formula, we get,

$x_{1,2}=\frac{-(-2)\pm\sqrt{(-2)^2-4\cdot (-3)\cdot 6} }{2(-3)}$

Simplify, we have,

$x_{1,2}=\frac{2\pm\sqrt{76} }{-6}$

Thus, $x_{1}=\frac{2+\sqrt{76} }{-6}$ and x_{2}=\frac{2-\sqrt{76} }{-6}

Simplify, we get,

$x_1=-\frac{1+\sqrt{19}}{3},\:x_2=\frac{\sqrt{19}-1}{3}$

Thus, $x_{1,2}=\frac{-(-2)\pm\sqrt{(-2)^2-4\cdot (-3)\cdot 6} }{2(-3)}$ shows correct substitution of the values a, b, and c from the given quadratic equation -3x^2-2x+6=0 into quadratic formula.

Answer from: ShelleyPSeliger

You can use the standard form of quadratic equation to find the correct substitution of the values a, b and c.

The correct substitution of the value a,b and c from the given quadratic equations into the solution form is $x = \dfrac{-(-2) \pm \sqrt{(-2)^2 - 4 \times -3 \times 6}}{2 \times (-3)}$

What is the standard form of quadratic equations?

The standard form of quadratic equations is ax^2 + bx + c = 0

What are the solutions of standard quadratic equation?

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

From given equation 0 = -3x^2 -2x + 6 , we deduce that a = -3, b = -2 and c = 6.

Thus, substituting these values in the solutions' equation:

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\
\\
x = \dfrac{2 \pm \sqrt{4 + 72}}{-6} = \dfrac{2\pm 2\sqrt{19}}{-6} = \dfrac{-1 \mp 2\sqrt{19}}{3}$

Thus, the correct substitution of the value a,b and c from the given quadratic equations into the solution form is $x = \dfrac{-(-2) \pm \sqrt{(-2)^2 - 4 \times -3 \times 6}}{2 \times (-3)}$

Learn more about solutions of quadratic equations here:

link

Answer from: Quest

their is no graph showing

step-by-step explanation:

Answer from: Quest

step-by-step explanation:

the 2000 florida legislature amended section 322.05, florida statutes, changing the requirements to obtain a class e license for a driver under the age of 18 holding a learner's license. the following requirements must be met in order to obtain a regular class e license if a learner's license is issued on or after october 1, 2000:

must hold a learner's license for at least 12 months or until the 18th birthday.

must have no convictions 12 months from the issue date of the learner's license.

may have one traffic conviction within 12 months from the issue date of the learner's license if adjudication is withheld.

a parent, legal guardian, or responsible adult over 21 years old must certify that the driver has 50 hours driving experience, including 10 hours of night time driving.