SOLUTION: Please help me solve these: 2x^2-x-28=0 x^2-2x=15 2x(x+7)(x+4)=0 x(x-5)=-6 x^2=16x

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Please help me solve these: 2x^2-x-28=0 x^2-2x=15 2x(x+7)(x+4)=0 x(x-5)=-6 x^2=16x      Log On


   

Question 535691: Please help me solve these:
2x^2-x-28=0
x^2-2x=15
2x(x+7)(x+4)=0
x(x-5)=-6
x^2=16x
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
There are four ways you can solve quadratic equations:
1) Use the quadratic formula: x+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a
2) Completing the square.
3) Factoring.
4) Use algebraic methods.
In problems 1 through 4, you can use factoring and in problem 5, use algebra:
1) 2x%5E2-x-28+=+0 Factor.
%282x%2B7%29%28x-4%29+=+0 Apply the zero product rule.
2x%2B7+=+0 or x-4+=+0
2x+=+-7 so that x+=+-7%2F2 or
x-4+=+0 so x+=+4
2) x%5E2-2x+=+15 Subtract 15 from both sides.
x%5E2-2x-15+=+0 Factor.
%28x%2B3%29%28x-5%29+=+0 Apply the zero product rule.
x%2B3+=+0 or x-5+=+0 so that...
x+=+-3 or x+=+5
3) 2x%28x%2B7%29%28x%2B4%29+=+0 This is already factored so apply the zero product rule.
2x+=+0 or x%2B7+=+0 or x%2B4+=+0 so that...
x+=+0 or x+=+-7 or x+=+-4
4)x%28x-5%29+=+-6 Expand the left side.
x%5E2-5x+=+-6 Add 6 to both sides.
x%5E2-5x%2B6+=+0 Factor.
%28x-2%29%28x-3%29+=+0 Apply the zero product rule.
x-2+=+0 or x-3+=+0 so that...
x+=+2 or x+=+3
5) x%5E2+=+16x Use algebra. Divide both sides by x.
x+=+16