Question

Solve the quadratic equation by finding square roots or by using the quadratic formula. Explain why you chose the method. $$6 x^{2}+20 x+5=0$$

Short answer

The solutions to the quadratic equation \(6x^{2}+20x+5=0\) are \(x_{1,2}= -\frac{10 \pm \sqrt{70}}{6}\)

Step-by-step solution

Write down the quadratic equation

The given quadratic equation is \(6x^{2}+20x+5=0\)

Identify a, b, c from the equation

From the given equation, we can identify the values of \(a\), \(b\), and \(c\) for the quadratic formula \(-\frac{b \pm \sqrt{b^{2}-4ac}}{2a}\) as \(a=6\), \(b=20\), and \(c=5\)

Apply the quadratic formula

Substitute the values \(a=6\), \(b=20\), and \(c=5\) into the quadratic formula \(-\frac{b \pm \sqrt{b^{2}-4ac}}{2a}\) to get the solutions

Compute the values under square root

Calculate the value under the square root \(b^{2}-4ac = (20)^{2}-4*6*5 = 400 -120 = 280\)

Apply the quadratic formula

Apply the quadratic formula with the values \(b=20\), \(a=6\), and \(\sqrt{280}\) to get the solutions. So, \(x_{1,2}= -\frac{20 \pm \sqrt{280}}{2*6}\)

Simplify

On simplifying the above equation, we get \(x_{1,2}= -\frac{10 \pm \sqrt{70}}{6}\)