Question

4 times a number is 21 less than the square of that number. Find the positive solution ​

Answer 1

set up the equation 4x + 21 = x^2. getting everything on one side so you have x^2-4x-21= 0. find two numbers that multiply to -21 and add to -4, which are 3 and -7. substitute in with them as x to get x^2+3x-7x+21 = 0. factor out the x and the 7 to get x(x+3)-7(x+3) = 21. so x = -3 and 7. the positive solution would be 7.