SOLUTION: Please help!!! I have tried solving the following problems, but can't seem to understand. Whatever assistance anyone can provide will be greatly appreciated. Thanks in advance.

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Please help!!! I have tried solving the following problems, but can't seem to understand. Whatever assistance anyone can provide will be greatly appreciated. Thanks in advance.       Log On


   



Question 76858This question is from textbook Beginning Algebra
: Please help!!! I have tried solving the following problems, but can't seem to understand. Whatever assistance anyone can provide will be greatly appreciated. Thanks in advance.
PG. 575
Using the factoring method, solve for the roots of each quadratic equation. Be sure to place your equation in standard form before factoring.
(14.)x^2 + 3x - 40 = 0
PG. 588
Solve using the quadratic formula. If there are no real roots, say so.
(20.) 3x^2 - 4x + 2 = 0
This question is from textbook Beginning Algebra

Answer by checkley75(3666) About Me  (Show Source):
You can put this solution on YOUR website!
X^2+3X-40=0
(X-5)(X+8)=0
X-5=0
X=5 ANSWER.
X+8=0
X=-8 ANSWER.
------------------------------------------------------------
3X^2-4X+2=0
USING THE QUADRATIC EQUATION WE GET:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
X=(4+-SQRT[-4*-4-4*3*2])/2*3
X=(4+-SQRT[16-24])/6
X=(4+-SQRT-8)/6 BECAUSE THERE IS A NEGATIVE NUMBER UNDER THE SQRT SIGN THE ANSWERS WILL NOT BE REAL ROOTS. THUS
X=(4+-2.83i)/6
X=4/6+2.83i/6
X=2/3+,47i ANSWER.
X=(4-2.38i)/6
X=4/6-2.38i/6
X=2/3-.47i ANSWER.