Hi,
Work:
Equation;
[tex]2 {x}^{2} + 7x - 15 = 0[/tex]
Solving:
[tex]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} [/tex]
Final solutions:
[tex]x1 = - 5 \\ x2 = \frac{3}{2} = 1 \frac{1}{2} = 1.5[/tex]
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
