Sure, let's go through the step-by-step solution for solving the quadratic equation:
Given equation:
[tex]\[ 11r + 15 = -2r^2 \][/tex]
First, let's rewrite it in standard form:
[tex]\[ 2r^2 + 11r + 15 = 0 \][/tex]
Now, we need to factorize the quadratic equation:
[tex]\[ 2r^2 + 11r + 15 = 0 \][/tex]
To factorize it, we look for two numbers that multiply to [tex]\( 2 \times 15 = 30 \)[/tex] and add to [tex]\( 11 \)[/tex].
The pair of numbers that satisfy these conditions is [tex]\( 6 \)[/tex] and [tex]\( 5 \)[/tex].
We can rewrite [tex]\( 11r \)[/tex] as [tex]\( 6r + 5r \)[/tex] to help us factor by grouping:
[tex]\[ 2r^2 + 6r + 5r + 15 = 0 \][/tex]
Next, we group the terms:
[tex]\[ (2r^2 + 6r) + (5r + 15) = 0 \][/tex]
Factor out the greatest common factor (GCF) from each group:
[tex]\[ 2r(r + 3) + 5(r + 3) = 0 \][/tex]
Now, we can factor out the common binomial factor [tex]\((r + 3)\)[/tex]:
[tex]\[ (r + 3)(2r + 5) = 0 \][/tex]
To find the solutions for [tex]\( r \)[/tex], we set each factor equal to zero:
[tex]\[ r + 3 = 0 \quad \text{or} \quad 2r + 5 = 0 \][/tex]
Solving these equations separately:
[tex]\[ r + 3 = 0 \Rightarrow r = -3 \][/tex]
[tex]\[ 2r + 5 = 0 \Rightarrow 2r = -5 \Rightarrow r = -\frac{5}{2} \][/tex]
Therefore, the solutions to the quadratic equation [tex]\( 2r^2 + 11r + 15 = 0 \)[/tex] are:
[tex]\[ r = -3 \][/tex]
[tex]\[ r = -\frac{5}{2} \][/tex]
