Question

Solve the equation for all real solutions in simplest form. 2d²+15d+23=3

Answer 1

To solve the quadratic equation 2d² + 15d + 23 = 3, subtract 3 from both sides, use the quadratic formula with a=2, b=15, c=20, and simplify to get two real solutions: d = (-15 + √65)/4 and d = (-15 - √65)/4.

Step-by-step explanation:

You are asking to solve the equation for all real solutions in simplest form: 2d² + 15d + 23 = 3. Let's move through the steps to solve this quadratic equation.

First, let's simplify the equation by subtracting 3 from both sides, which gives us:

2d² + 15d + 20 = 0.

Now, we can use the quadratic formula to solve for d. The quadratic formula is d = (-b ± √(b² - 4ac))/(2a) for an equation of the form ax² + bx + c = 0. In our case, a = 2, b = 15, and c = 20.

Substituting these values into the quadratic formula gives us two potential solutions for d:

d = (-15 ± √(15² - 4×2×20))/(2×2).

Calculating further, we get:

d = (-15 ± √(225 - 160))/4,
which simplifies to:

d = (-15 ± √65)/4.

Therefore, the two real solutions in simplest form are:

d = (-15 + √65)/4 and d = (-15 - √65)/4.

Answer 2

To solve the equation \(2d^2 + 15d + 23 = 3\), let's first simplify the equation:

\(2d^2 + 15d + 23 = 3\)

Subtract 3 from both sides:

\(2d^2 + 15d + 20 = 0\)

Now, to solve for \(d\), you can either use the quadratic formula or factor the equation.

Let's use the quadratic formula: \(d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = 15\), and \(c = 20\).

Plugging these values into the quadratic formula:

\(d = \frac{-15 \pm \sqrt{15^2 - 4(2)(20)}}{2(2)}\)

\(d = \frac{-15 \pm \sqrt{225 - 160}}{4}\)

\(d = \frac{-15 \pm \sqrt{65}}{4}\)

So, the solutions for \(d\) are:

\(d = \frac{-15 + \sqrt{65}}{4}\) and \(d = \frac{-15 - \sqrt{65}}{4}\)