Solving quadratic equations with the quadratic formula: 3 worked examples

To solve the quadratic equation $2x^2 + 3x - 5 = 0$ using the quadratic formula, we identify the coefficients $a$, $b$, and $c$ from the standard form $ax^2 + bx + c = 0$. Here, $a = 2$, $b = 3$, and $c = -5$.

The quadratic formula is given by:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting the values into the formula:
$x = \frac{-(3) \pm \sqrt{(3)^2 - 4(2)(-5)}}{2(2)}$

Next, simplify the expression under the square root (the discriminant):
$x = \frac{-3 \pm \sqrt{9 - (-40)}}{4}$
$x = \frac{-3 \pm \sqrt{9 + 40}}{4}$
$x = \frac{-3 \pm \sqrt{49}}{4}$

Since $\sqrt{49} = 7$:
$x = \frac{-3 \pm 7}{4}$

This gives us two possible values for $x$:

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 $x = 1$ and $x = -\frac{5}{2}$.

To solve the quadratic equation $x^{2} - 6x + 4 = 0$ using the quadratic formula, we identify the coefficients $a$, $b$, and $c$ from the standard form $ax^{2} + bx + c = 0$. Here, $a = 1$, $b = -6$, and $c = 4$.

The quadratic formula is given by:
$x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$

Substituting the identified values into the formula:
$x = \frac{-(-6) \pm \sqrt{(-6)^{2} - 4(1)(4)}}{2(1)}$
$x = \frac{6 \pm \sqrt{36 - 16}}{2}$
$x = \frac{6 \pm \sqrt{20}}{2}$

Next, we simplify $\sqrt{20}$. Since $20 = 4 \times 5$, we have $\sqrt{20} = \sqrt{4} \times \sqrt{5} = 2\sqrt{5}$.
$x = \frac{6 \pm 2\sqrt{5}}{2}$

Dividing both terms in the numerator by the denominator $2$:
$x = \frac{6}{2} \pm \frac{2\sqrt{5}}{2}$
$x = 3 \pm \sqrt{5}$

To solve the quadratic equation $3x^{2} + 2x - 1 = 0$ using the quadratic formula, we identify the coefficients: $a = 3$, $b = 2$, and $c = -1$. The quadratic formula is given by $x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$.

First, calculate the discriminant $D = b^{2} - 4ac$:
$D = (2)^{2} - 4(3)(-1)$
$D = 4 - (-12)$
$D = 4 + 12 = 16$

Next, substitute the values into the formula:
$x = \frac{-2 \pm \sqrt{16}}{2(3)}$
$x = \frac{-2 \pm 4}{6}$

This gives two possible solutions:

1. $x = \frac{-2 + 4}{6} = \frac{2}{6} = \frac{1}{3}$
2. $x = \frac{-2 - 4}{6} = \frac{-6}{6} = -1$