Solve for z z^2-3/4z=-80

$z^{2} - \frac{3}{4} z = - 80$

Step 1

Simplify each term.

Combine $z$ and $\frac{3}{4}$ .

$z^{2} - \frac{z \cdot 3}{4} = - 80$

Move $3$ to the left of $z$ .

$z^{2} - \frac{3 z}{4} = - 80$

Step 2

Add $80$ to both sides of the equation.

$z^{2} - \frac{3 z}{4} + 80 = 0$

Step 3

Multiply through by the least common denominator $4$ , then simplify.

Apply the distributive property.

$4 z^{2} + 4 (- \frac{3 z}{4}) + 4 \cdot 80 = 0$

Simplify.

Cancel the common factor of $4$ .

Move the leading negative in $- \frac{3 z}{4}$ into the numerator.

$4 z^{2} + 4 (\frac{- 3 z}{4}) + 4 \cdot 80 = 0$

Cancel the common factor.

$4 z^{2} + 4 (\frac{- 3 z}{4}) + 4 \cdot 80 = 0$

Rewrite the expression.

$4 z^{2} - 3 z + 4 \cdot 80 = 0$

Multiply $4$ by $80$ .

$4 z^{2} - 3 z + 320 = 0$

Step 4

Use the quadratic formula to find the solutions.

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

Step 5

Substitute the values $a = 4$ , $b = - 3$ , and $c = 320$ into the quadratic formula and solve for $z$ .

$\frac{3 \pm \sqrt{{(- 3)}^{2} - 4 \cdot (4 \cdot 320)}}{2 \cdot 4}$

Step 6

Simplify.

Simplify the numerator.

Raise $- 3$ to the power of $2$ .

$z = \frac{3 \pm \sqrt{9 - 4 \cdot 4 \cdot 320}}{2 \cdot 4}$

Multiply $- 4 \cdot 4 \cdot 320$ .

Multiply $- 4$ by $4$ .

$z = \frac{3 \pm \sqrt{9 - 16 \cdot 320}}{2 \cdot 4}$

Multiply $- 16$ by $320$ .

$z = \frac{3 \pm \sqrt{9 - 5120}}{2 \cdot 4}$

Subtract $5120$ from $9$ .

$z = \frac{3 \pm \sqrt{- 5111}}{2 \cdot 4}$

Rewrite $- 5111$ as $- 1 (5111)$ .

$z = \frac{3 \pm \sqrt{- 1 \cdot 5111}}{2 \cdot 4}$

Rewrite $\sqrt{- 1 (5111)}$ as $\sqrt{- 1} \cdot \sqrt{5111}$ .

$z = \frac{3 \pm \sqrt{- 1} \cdot \sqrt{5111}}{2 \cdot 4}$

Rewrite $\sqrt{- 1}$ as $i$ .

$z = \frac{3 \pm i \sqrt{5111}}{2 \cdot 4}$

Multiply $2$ by $4$ .

$z = \frac{3 \pm i \sqrt{5111}}{8}$

Step 7

The final answer is the combination of both solutions.

$z = \frac{3 + i \sqrt{5111}}{8}, \frac{3 - i \sqrt{5111}}{8}$

Solve for z z^2-3/4z=-80

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