Question

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use the quadratic formula to solve. express your answer in simplest form.
$-3r^2 + 9r + 4 = -5r^2$
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attempt 1 out of 4
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Explanation:

Step1: Rearrange the equation

First, we need to get the quadratic equation in standard form \(ax^{2}+bx + c = 0\). Starting with \(-3r^{2}+9r + 4=-5r^{2}\), we add \(5r^{2}\) to both sides:
\(-3r^{2}+5r^{2}+9r + 4=0\)
Simplify the \(r^{2}\) terms: \(2r^{2}+9r + 4 = 0\)

Step2: Identify \(a\), \(b\), \(c\)

For the quadratic equation \(2r^{2}+9r + 4 = 0\), we have \(a = 2\), \(b = 9\), \(c = 4\)

Step3: Apply the quadratic formula

The quadratic formula is \(r=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Substitute \(a = 2\), \(b = 9\), \(c = 4\) into the formula:
First, calculate the discriminant \(\Delta=b^{2}-4ac=(9)^{2}-4\times2\times4=81 - 32 = 49\)
Then, \(r=\frac{-9\pm\sqrt{49}}{2\times2}=\frac{-9\pm7}{4}\)

Step4: Find the two solutions

For the plus sign: \(r=\frac{-9 + 7}{4}=\frac{-2}{4}=-\frac{1}{2}\)
For the minus sign: \(r=\frac{-9-7}{4}=\frac{-16}{4}=-4\)

Answer:

\(r = -\frac{1}{2}\) or \(r=-4\)