Respuesta :
The formula for the solution is:
[tex]x=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
In your case: a=-2, b=3, c=5
Replace these in the formula and you'll get:
[tex]x_1=\frac{-b + \sqrt{b^2 - 4ac}}{2a}=\frac{-3 + \sqrt{3^2 - 4 \cdot (-2) \cdot 5}}{2 \cdot (-2)}=-1[/tex]
[tex]x_2=\frac{-b - \sqrt{b^2 - 4ac}}{2a}=\frac{-3 - \sqrt{3^2 - 4 \cdot (-2) \cdot 5}}{2 \cdot (-2)}= \frac{5}{2}[/tex]
So you have two roots:
x1 = -1
x2 = 5/2
[tex]x=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
In your case: a=-2, b=3, c=5
Replace these in the formula and you'll get:
[tex]x_1=\frac{-b + \sqrt{b^2 - 4ac}}{2a}=\frac{-3 + \sqrt{3^2 - 4 \cdot (-2) \cdot 5}}{2 \cdot (-2)}=-1[/tex]
[tex]x_2=\frac{-b - \sqrt{b^2 - 4ac}}{2a}=\frac{-3 - \sqrt{3^2 - 4 \cdot (-2) \cdot 5}}{2 \cdot (-2)}= \frac{5}{2}[/tex]
So you have two roots:
x1 = -1
x2 = 5/2