Michael J. answered • 06/09/15

Effective High School STEM Tutor & CUNY Math Peer Leader

A quadratic equation is always in the form

ax

^{2}+ bx + c = 0where:

a, b, and c are constants.

The quadratic formula is

x = (-b ± √(b

^{2}- 4ac)) / 2aThe plus/minus sign indicates that you will have two solutions.

From the first part of the problem, if we subtract 14x on both sides of equation, we obtain

5x

^{2}- 14x - 3 = 0a = 5

b = -14

c = -3

From the second part of the problem, if we subtract x and subtract 1/6 on both sides of equation, we obtain.

(1/3)x

^{2}- x - (1/6) = 0a = 1/3

b = -1

c = -1/6

Plug in the values of a, b, and c into the formula for each equation respectively to obtain your solutions.

Andrew M.

on the 2nd problem:

We can actually use 6 instead of 18 to eliminate the fraction... the smaller numbers would be obtained by multiplying the original equation through by 6 giving 2x

^{2}-6x-1This gives a = 2, b = -6, c = -1

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Also, David... you accidentally switched the sign of the constant... your equation should be

6x

^{2}-18x - 3I just reduced it again by dividing out the 3 as a common factor of that

Report

06/10/15

David W.

^{2 }- 18x + 3 = 006/09/15