# Algebra

How many real number solutions does the equation have? 0 = -7x^2 + 6x + 3
one solution
two solutions**
no solution

How many real number solutions does the equation have? 0 = 2x^2 - 20x + 50
one solution
two solutions
no solution**

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1. #1 ok
#2 nope. the polynomial is 2(x-5)^2
always check the discriminant. For this one, it is zero, indicating one repeated root.

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2. Find the discriminant: D=b^2-4ac

if D>0, there are two real number solutions
if D=0, there is one solution
if D<0, there are no solutions

Using the quadratic equation: ax^2 + bx +c, insert the values into the discriminant.

1) -7x^2 + 6x + 3

D=(6)^2 - 4(-7)(3)
=36+84
=120>0
Therefore, there are 2 real number solutions.

2) 2x^2 - 20x + 50

D=400-400
= 0
Therefore, there is one solution.

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3. For the second one I thought it was negative?
(-20)^2 - 4(2)(50) =
- 400 - 400 = -800?

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4. sorry ... b^2 = (-20)^2 = (-20)(-20) = 400 not -400
-400 is -20^2 = -b^2 because exponents are done first.

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5. thanks soo much everyone!! :3

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