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Created on: 2012-03-20

A sample problem solved by Quickmath algebra solver

Command

Solve

Equation
-t^2+4*t-8 = 0

  1. an equation LHS is equal to a sum consisting of 3 terms; the first term of the sum is a negative power; the base is t; the exponent is two; the second term of the sum is equal to a product comprising 2 factors; the first factor of the product is equal to four; the second factor of the product is t; the third term of the sum is equal to negative eight Right side of the equation is equal to zero;
  2. negative t to the power of two plus four multiplied by t plus negative eight is equal to zero.
Variable
Result

Exact

Solution 1 ( complex )
t = 2-2*%i

  1. an equation LHS is t RHS is equal to a sum containing 2 terms. The first term of the sum is equal to two. The second term of the sum is a negative product consisting of 2 factors. The first factor of the product is equal to two. The second factor of the product is equal to i.
  2. t is two plus negative two times i.
Solution 2 ( complex )
t = 2*%i+2

  1. an equation LHS is equal to t RHS is a sum that contains 2 terms. The first term of the sum is equal to a product containing 2 factors. The first factor of the product is two. The second factor of the product is equal to i. The second term of the sum is two.
  2. t is equal to two times i plus two.