Question

question
solve for the roots in simplest form using the quadratic formula:
$4x^2 + 53 = -28x$

answer attempt 2 out of 2

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Explanation:

Step1: Rewrite in standard form

First, we rewrite the equation \(4x^{2}+53 = - 28x\) into the standard quadratic form \(ax^{2}+bx + c=0\).
Adding \(28x\) to both sides, we get \(4x^{2}+28x + 53=0\). Here, \(a = 4\), \(b=28\), and \(c = 53\).

Step2: Apply quadratic formula

The quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
First, calculate the discriminant \(\Delta=b^{2}-4ac\). Substitute \(a = 4\), \(b = 28\), and \(c=53\) into the discriminant formula:
\(\Delta=(28)^{2}-4\times4\times53\)
\(=784 - 848\)
\(=- 64\)

Since the discriminant \(\Delta=- 64<0\), and in the real - number system, the square root of a negative number is not defined. So, the equation \(4x^{2}+28x + 53 = 0\) (which is equivalent to the original equation \(4x^{2}+53=-28x\)) has no real solutions.

Answer:

No solution