Alegbra II help? Pleaaase

Find the discriminate of the quadratic.
Identify whether there are;
2 Real Solutions
1 Real Solution
2 Imaginary Solutions


1. 8x*2 - 9n + 13 = 0



Use any method to solve the following equations:

1. 3x^2 + 7x - 24 = 13x
2. x^2 - 6x + 7 = 0



Thank you, guys.

8x² - 9x + 13 = 0. This is in standard ax² + bx + c = 0 form, where a = 8, b = -9,
and c = 15.
The discriminant, form the quadratic formula, is b² - 4ac, which here is (-9)² - 4(8)(13)
= 81 - 416
= -335.

Since -335 < 0, the equation has 2 imaginary solutions.
(If the discriminant is > 0, there are 2 real solutions; if it = 0, there is 1 real solution --- a "repeated root".)


3x² + 7x - 24 = 13x.
Subtract 13x from both sides, giving
3x² - 6x - 24 = 0.
Divide everything by 3, for simplicity, giving
x² - 2x - 8 = 0.
Factor this as
(x+2)(x-4) = 0, so x = -2 or x = 4.


x² - 6x + 7 = 0.
Factor this as
(x+1)(x-7) = 0, so x = -1 or x = 7.

(x+1)(x-7) is x^2 -6x -7 not +7

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given a quadratic in the form of ax^2 + bx +c then the discriminate is b^2 - 4ac

(-9)^2 - 4*8*13 = -335

as the discriminate is less than zero there are 2 Imaginary Solutions


3x^2 + 7x - 24 = 13x
=> 3x^2 -6x -24 = 0
=> 3x^2 -6x = 24
=> x^2 -2x = 8
=> x^2 -2x + 1 = 9
=> (x - 1)^2 = 9
=> x - 1 = ±3
=> x = 1±3
=> x = -2 or 4


x^2 - 6x + 7 = 0
=> x^2 -6x = -7
=> x^2 -6x + 9 = 2
=> (x -3)^2 = 2
=> x -3 = ±√2
=> x = 3±√2
=> x = 3-√2 and 3+√2