Find All Complex Solutions 20x^2-47x+24=0

$20 x^{2} - 47 x + 24 = 0$

Step 1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2

Substitute the values $a = 20$ , $b = - 47$ , and $c = 24$ into the quadratic formula and solve for $x$ .

$\frac{47 \pm \sqrt{{(- 47)}^{2} - 4 \cdot (20 \cdot 24)}}{2 \cdot 20}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 47$ to the power of $2$ .

$x = \frac{47 \pm \sqrt{2209 - 4 \cdot 20 \cdot 24}}{2 \cdot 20}$

Multiply $- 4 \cdot 20 \cdot 24$ .

Multiply $- 4$ by $20$ .

$x = \frac{47 \pm \sqrt{2209 - 80 \cdot 24}}{2 \cdot 20}$

Multiply $- 80$ by $24$ .

$x = \frac{47 \pm \sqrt{2209 - 1920}}{2 \cdot 20}$

Subtract $1920$ from $2209$ .

$x = \frac{47 \pm \sqrt{289}}{2 \cdot 20}$

Rewrite $289$ as $17^{2}$ .

$x = \frac{47 \pm \sqrt{17^{2}}}{2 \cdot 20}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{47 \pm 17}{2 \cdot 20}$

Multiply $2$ by $20$ .

$x = \frac{47 \pm 17}{40}$

Step 4

The final answer is the combination of both solutions.

$x = \frac{8}{5}, \frac{3}{4}$