The given equation is a quadratic equation in the form of ax^2 + bx + c = 0. The general solution for such equations is given by the quadratic formula: x = [-b ± sqrt(b^2 - 4ac)] / (2a).

Here, a = 1 (coefficient of d^2), b = 19 (coefficient of d), and c = -20.

Let's substitute these values into the quadratic formula:

d = [-19 ± sqrt((19)^2 - 4*1*(-20))] / (2*1)
d = [-19 ± sqrt(361 + 80)] / 2
d = [-19 ± sqrt(441)] / 2
d = [-19 ± 21] / 2

This gives us two solutions:

d = (-19 + 21) / 2 = 2 / 2 = 1
d = (-19 - 21) / 2 = -40 / 2 = -20

So, the solutions to the equation d^2 + 19d - 20 = 0 are d = 1 and d = -20.

Question

The given equation is a quadratic equation in the form of ax^2 + bx + c = 0. The general solution for such equations is given by the quadratic formula: x = [-b ± sqrt(b^2 - 4ac)] / (2a).

Here, a = 1 (coefficient of d^2), b = 19 (coefficient of d), and c = -20.

Let's substitute these values into the quadratic formula:

d = [-19 ± sqrt((19)^2 - 4*1*(-20))] / (2*1)
d = [-19 ± sqrt(361 + 80)] / 2
d = [-19 ± sqrt(441)] / 2
d = [-19 ± 21] / 2

This gives us two solutions:

d = (-19 + 21) / 2 = 2 / 2 = 1
d = (-19 - 21) / 2 = -40 / 2 = -20

So, the solutions to the equation d^2 + 19d - 20 = 0 are d = 1 and d = -20.

Knowee AI · Accepted Answer

The given equation is a quadratic equation in the form of ax^2 + bx + c = 0. The general solution for such equations is given by the quadratic formula: x = [-b ± sqrt(b^2 - 4ac)] / (2a).

Here, a = 1 (coefficient of d^2), b = 19 (coefficient of d), and c = -20.

Let's substitute these values into the quadratic formula:

d = [-19 ± sqrt((19)^2 - 4*1*(-20))] / (2*1)
d = [-19 ± sqrt(361 + 80)] / 2
d = [-19 ± sqrt(441)] / 2
d = [-19 ± 21] / 2

This gives us two solutions:

d = (-19 + 21) / 2 = 2 / 2 = 1
d = (-19 - 21) / 2 = -40 / 2 = -20

So, the solutions to the equation d^2 + 19d - 20 = 0 are d = 1 and d = -20.

Solve for d.d2+19d–20=0Write each solution as an integer, proper fraction, or improper fraction in simplest form. If there are multiple solutions, separate them with commas.

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