How do you solve the equation $ 2x^2 + 3x - 5 = 0 $ using the quadratic formula?
Answer 1
Kelly Leblanc
To solve the quadratic equation $2x^2 + 3x – 5 = 0$ using the quadratic formula, we use the formula:
$$ x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} $$
Here, $a = 2$, $b = 3$, and $c = -5$. Plugging in these values, we get:
$$ x = \frac{-3 \pm \sqrt{3^2 – 4 \cdot 2 \cdot (-5)}}{2 \cdot 2} $$
Simplifying the expression:
$$ x = \frac{-3 \pm \sqrt{9 + 40}}{4} $$
$$ x = \frac{-3 \pm \sqrt{49}}{4} $$
Therefore, we have:
$$ x = \frac{-3 \pm 7}{4} $$
Thus, the two solutions are:
$$ x = \frac{-3 + 7}{4} = \frac{4}{4} = 1 $$
and
$$ x = \frac{-3 – 7}{4} = \frac{-10}{4} = -\frac{5}{2} $$
Answer 2
Maria Rodriguez
To solve the quadratic equation $2x^2 + 3x – 5 = 0$ using the quadratic formula:
$$ x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} $$
For $a = 2$, $b = 3$, and $c = -5$:
$$ x = \frac{-3 \pm \sqrt{9 + 40}}{4} $$
$$ x = \frac{-3 \pm \sqrt{49}}{4} $$
So, the two solutions are:
$$ x = \frac{4}{4} = 1 $$
and
$$ x = \frac{-10}{4} = -\frac{5}{2} $$
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