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Answer :

XKelvingo

Answer:

C

Step-by-step explanation:

We have the equation:

[tex]4x^2+5x=-10[/tex]

Add 10 to both sides to isolate the equation.

[tex]4x^2+5x+10=0[/tex]

This is not factorable*, so we can use the quadratic formula:

[tex]\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]

In this case, a = 4, b = 5, and c = 10.

Substitute:

[tex]\displaystyle x=\frac{-(5)\pm\sqrt{(5)^2-4(4)(10)}}{2(4)}[/tex]

Simplify:

[tex]\displaystyle x=\frac{-5\pm\sqrt{-135}}{8}[/tex]

Since we cannot take the root of a negative, we have no real solutions.

Our answer is C.

*To factor something in the form of:

[tex]ax^2+bx+c=0[/tex]

We want two numbers p and q such that pq = ac and p + q = b.

Since ac = 4(10) = 40. We need to find two whole numbers that multiply to 40 and add to 5.

No such numbers exist, so the equation is not factorable.

Miraculouslbcomicsgo

Answer:

c) No Real Solution

Step-by-step explanation:

I'ma try my best to explain it!

We have: 4x^2+5x=−10

Step 1: Subtract -10 from both sides.

4x^2+5x−(−10)=−10−(−10) (you don't put the -10 next to it but you put it in the bottom)

Then we get 4x^2+5x+10=0 from the subtraction above

For this equation: a=4, b=5, c=10

4x^2+5x+10=0

Then we use quadratic formula with a=4, b=5, c=10

The answer is the picture below: means no solution

Hope this helps and is correct!

View image Miraculouslbcomics

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