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.
