Solve 2x^2 + 7x = -8 using the Quadratic Formula.

A: No solution?

2x^2 + 7x + 8 = 0

B^2 = 7^2 = 49
4AC = 4*2*8 = 64.

Since B^2 is less than 4AC, there are
no real solution. But there are 2 imaginary solutions:

X = (-B +-sqrt(B^2-4AC))/2A
X =(-7 +-sqrt(49-64))/4
X = (-7+-sqrt(-15))/4
X = (-7+-sqrt(15*-1))/4
X = (-7+-(3.87i))/4

X = -1.75 + 0.968i, and -1.75 - 0.968i.

For 2 real solutions: B^2 => 4AC

To solve the quadratic equation 2x^2 + 7x = -8 using the quadratic formula, we first need to rearrange the equation into the standard form: ax^2 + bx + c = 0. In this case, a = 2, b = 7, and c = 8.

The quadratic formula is given by:

x = (-b ± √(b^2 - 4ac))/(2a)

Substituting the values of a, b, and c into the formula, we get:

x = (-(7) ± √((7)^2 - 4(2)(-8)))/(2(2))

Simplifying further:

x = (-7 ± √(49+64))/4

x = (-7 ± √(113))/4

Since the quadratic formula involves the square root of a number, we need to determine whether the expression inside the square root is positive, negative, or zero. In this case, the expression 113 is positive, so we can continue.

Therefore, the solutions to the equation 2x^2 + 7x = -8 are:

x = (-7 + √113)/4 and x = (-7 - √113)/4

So, there are two solutions to this equation.