Question 1076100
Please help me solve this by COMPLETING THE SQUARE
28x-2x^2=26

I got no real solution. This is what I did, please correct me.

-2x^2+28x=26
(28/2)^2=196
-2x^2+28x+196=26+196
since you cannot take the square root of a negative, i got no real solution
 
√-2x^2 = no real solutions

Please help me, I am very confused. Thank you in advance, Emily
<pre>First of all, you're INCORRECT. You should see that after FACTORING OUT a GCF,  2, you'll end up with {{{y = x^2 - 14x + 13}}}.
This can be FACTORED to get roots/solutions/zeroes of x = 13, or x = 1.
From this, it's clear that these will be your solutions, and they are REAL.
You could have also used the DISCRIMINANT to see if it really has No real solutions. If you did. You'd see that it does have 2 REAL solutions.

Now that we know this, let's get to the solutions by COMPLETING the SQUARE.
 {{{28x - 2x^2 = 26}}}
{{{matrix(1,3, - 2(- 14x + x^2) = - 2(- 13), "======>", - 14x + x^2 = - 13)}}}
{{{x^2 - 14x = - 13}}} ------- Rearranging equation
{{{x^2 - 14x + ((1/2) * - 14)^2 = - 13 + ((1/2) * - 14)^2}}} -------- Taking {{{matrix(1,3, 1/2, of, b)}}}, squaring it, and adding the result to both sides
{{{x^2 - 14x + (- 7)^2 = - 13 + (- 7)^2}}}
{{{(x - 7)^2 = - 13 + 49}}}
{{{(x - 7)^2 = 36}}}
{{{sqrt((x - 7)^2) = " "+- sqrt(36)}}} ------ Taking the square root of both sides
{{{x - 7 = " "+- 6}}} ======> {{{x = " "+- 6 + 7}}}
{{{highlight_green(matrix(1,15, x, "=", 6 + 7, "=====>", x, "=", 13, OR, x, "=", - 6 + 7, "=====>", x, "=", 1)))}}}