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solve 2x^2+3x-5=0

ax^2+bx+c=0
a=___ b=___ c=___
x= -___± √__^2 - 4___ ___/2___
simplify -b= ___ b^2- 4ac = ___ 2a = ___
replace x= __ ± √___/___
x= ___ ___

1 answer

To solve the quadratic equation \(2x^2 + 3x - 5 = 0\), we can identify the coefficients:

  • \(a = 2\)
  • \(b = 3\)
  • \(c = -5\)

Now, we will apply the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Let's fill in the specific values:

  1. \(-b = -3\)
  2. \(b^2 - 4ac = 3^2 - 4 \cdot 2 \cdot (-5) = 9 + 40 = 49\)
  3. \(2a = 2 \cdot 2 = 4\)

Now we can insert these into the formula:

\[ x = \frac{-3 \pm \sqrt{49}}{4} \]

Since \(\sqrt{49} = 7\), we can complete the equation as follows:

\[ x = \frac{-3 \pm 7}{4} \]

This gives us two possible solutions:

  1. \(x = \frac{-3 + 7}{4} = \frac{4}{4} = 1\)
  2. \(x = \frac{-3 - 7}{4} = \frac{-10}{4} = -\frac{5}{2}\)

Thus, the solutions are:

\[ x = 1 \quad \text{and} \quad x = -\frac{5}{2} \]

Summing up, we have:

  • \(a = 2\)
  • \(b = 3\)
  • \(c = -5\)

The quadratic formula’s final solutions are:

\[ x = 1 \quad \text{or} \quad x = -\frac{5}{2} \]