Question

4x^2 + 3x = -6

Solve each equation by using the Quadratic Formula

Answer 1

add6 on both side

Answer 2

which side would you add the 6 too

Answer 3

First simplify it:

\(4x^2 + 3x = -6\)

Add 6 to both sides:

\(4x^2 + 3x + 6 = 0\)

Now it's in the form of \(ax^2 + bx + c\)

We can plug it into the Quadratic formula and solve:

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

Here:

\(a = 4\)
\(b = 3\)
\(c = 6\)

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

\(x = \dfrac{-3 \pm \sqrt{3^2 - 4(4)(6)}}{2(4)}\)

Multiply 2 * 4:

\(x = \dfrac{-3 \pm \sqrt{3^2 - 4(4)(6)}}{8}\)

Simplify exponent:

\(x = \dfrac{-3 \pm \sqrt{9 - 4(4)(6)}}{8}\)

Multiply:

\(x = \dfrac{-3 \pm \sqrt{9 - 16(6)}}{8}\)

Multiply:

\(x = \dfrac{-3 \pm \sqrt{9 - 96}}{8}\)

Subtract 9 - 96, what do you get?

Answer 4

@good_ole_southerngirl

Answer 5

Can you subtract that? @good_ole_southerngirl

Answer 6

-84

Answer 7

** -87

Answer 8

Yes, that gives us:

\(x = \dfrac{-3 \pm \sqrt{-87}}{8}\)

And since we can't take the square root of a negative number, there is no solutions.

Answer 9

Welcome to Open Study!

You can give medals by clicking 'Best Response'. @good_ole_southerngirl

Answer 10

Would you be able to help me with possibly 2 or 4 more?

Answer 11

I don't know if I did this right