Question

Using the discriminant , how many real solutions does the following quadratic equation have?

Answer 1

2x²+7x-15=0

a=2; b=7; c=-15

Δ=b²-4ac=49+120=169>0  =>

B. Two real solutions

x1=(-b+√Δ)/2a=(-7+13)/4=6/4=3/2

x2=(-b-√Δ)/2a=(-7-13)/4=-5

Answer 2

Hi,

Work:

Equation;

$2 {x}^{2} + 7x - 15 = 0$

Solving:

$2x^{2} + 7x - 15 = 0 \\ 2 {x}^{2} + 10x - 3x - 15 = 0 \\ 2x \times (x + 5) - 3x - 15 = 0 \\ 2x \times (x + 5) - 3(x + 5) = 0 \\ (x + 5) \times (2x - 3) = 0 \\ \\ x + 5 = 0 \\ 2x - 3 = 0 \\ \\ x = - 5 \\ x = \frac{3}{2}$

Final solutions:

$x1 = - 5 \\ x2 = \frac{3}{2} = 1 \frac{1}{2} = 1.5$

The answer is: Quadric equation has two real solutions.

Hope this helps.
If you need any additional explanation comment below and I'll reply.
r3t40