I need to know why these equations have solutions?

Question

5x^2+8x+7=0
(7)^1/2y^2-6y-13(7)^1/2=0
2x^2+x-1=0
4/3x^2-2x+3/4=0
Please explain to me how to get these solutions

Answer 1

5x^2+8x+7=0
This equation does not have proper integer solutions for x, but being a quadratic equation of the form ax^2 + bx + c = 0 has roots x1 = [-b + sqrt(b^2 - 4ac)] / 2a and x2 = [-b - sqrt(b^2 - 4ac)] / 2a

Similarly, other equations may either have simple factors or be solvable using the quadratic formula.

Answer 2

One way to determine whether quadratic equations have real solutions is checking its discriminant, b^2 - 4ac.

If b^2 - 4ac is greater than or equal to 0, it will have real solutions.

Specifically, if b^2 - 4ac = 0, it will have one (unique) solution, and if b^2 - 4ac > 0, it will have two real solutions.

Applying this to the first quadratic,

5x^2 + 8x + 7 = 0

a = 5, b = 8, c = 7

b^2 - 4ac = 8^2 - 4(5)(7)
= 64 - 140
= {some negative number}

So this equation has no real solutions. To get the complex solutions, just use the quadratic formula.

Answer 3

They have solutions because they are solvable equations.