The given equation is a quadratic equation in the form of ax^2 + bx + c = 0, where a = 1, b = 29, and c = 0.

The solutions to a quadratic equation are given by the formula x = [-b ± sqrt(b^2 - 4ac)] / 2a.

Substituting the values of a, b, and c into the formula, we get:

j = [-29 ± sqrt((29)^2 - 4*1*0)] / 2*1
j = [-29 ± sqrt(841)] / 2
j = [-29 ± 29] / 2

This gives us two solutions:

j = (0 / 2) = 0
j = (-58 / 2) = -29

So, the solutions to the equation are j = 0 and j = -29.

Question

The given equation is a quadratic equation in the form of ax^2 + bx + c = 0, where a = 1, b = 29, and c = 0.

The solutions to a quadratic equation are given by the formula x = [-b ± sqrt(b^2 - 4ac)] / 2a.

Substituting the values of a, b, and c into the formula, we get:

j = [-29 ± sqrt((29)^2 - 4*1*0)] / 2*1
j = [-29 ± sqrt(841)] / 2
j = [-29 ± 29] / 2

This gives us two solutions:

j = (0 / 2) = 0
j = (-58 / 2) = -29

So, the solutions to the equation are j = 0 and j = -29.

Knowee AI · Accepted Answer

The given equation is a quadratic equation in the form of ax^2 + bx + c = 0, where a = 1, b = 29, and c = 0.

The solutions to a quadratic equation are given by the formula x = [-b ± sqrt(b^2 - 4ac)] / 2a.

Substituting the values of a, b, and c into the formula, we get:

j = [-29 ± sqrt((29)^2 - 4*1*0)] / 2*1
j = [-29 ± sqrt(841)] / 2
j = [-29 ± 29] / 2

This gives us two solutions:

j = (0 / 2) = 0
j = (-58 / 2) = -29

So, the solutions to the equation are j = 0 and j = -29.

Solve for j.j2+29j=0Write each solution as an integer, proper fraction, or improper fraction in simplest form. If there are multiple solutions, separate them with commas.