Pre-Algebra Examples

$(- 3) = 4 x^{2} - 5 x$

Step 1

Rewrite the equation as $4 x^{2} - 5 x = - 3$ .

$4 x^{2} - 5 x = - 3$

Step 2

Add $3$ to both sides of the equation.

$4 x^{2} - 5 x + 3 = 0$

Step 3

Use the quadratic formula to find the solutions.

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

Step 4

Substitute the values $a = 4$ , $b = - 5$ , and $c = 3$ into the quadratic formula and solve for $x$ .

$\frac{5 \pm \sqrt{{(- 5)}^{2} - 4 \cdot (4 \cdot 3)}}{2 \cdot 4}$

Step 5

Simplify.

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Step 5.1

Simplify the numerator.

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Step 5.1.1

Raise $- 5$ to the power of $2$ .

$x = \frac{5 \pm \sqrt{25 - 4 \cdot 4 \cdot 3}}{2 \cdot 4}$

Step 5.1.2

Multiply $- 4 \cdot 4 \cdot 3$ .

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Step 5.1.2.1

Multiply $- 4$ by $4$ .

$x = \frac{5 \pm \sqrt{25 - 16 \cdot 3}}{2 \cdot 4}$

Step 5.1.2.2

Multiply $- 16$ by $3$ .

$x = \frac{5 \pm \sqrt{25 - 48}}{2 \cdot 4}$

Step 5.1.3

Subtract $48$ from $25$ .

$x = \frac{5 \pm \sqrt{- 23}}{2 \cdot 4}$

Step 5.1.4

Rewrite $- 23$ as $- 1 (23)$ .

$x = \frac{5 \pm \sqrt{- 1 \cdot 23}}{2 \cdot 4}$

Step 5.1.5

Rewrite $\sqrt{- 1 (23)}$ as $\sqrt{- 1} \cdot \sqrt{23}$ .

$x = \frac{5 \pm \sqrt{- 1} \cdot \sqrt{23}}{2 \cdot 4}$

Step 5.1.6

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

$x = \frac{5 \pm i \sqrt{23}}{2 \cdot 4}$

Step 5.2

Multiply $2$ by $4$ .

$x = \frac{5 \pm i \sqrt{23}}{8}$

Step 6

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 6.1

Simplify the numerator.

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Step 6.1.1

Raise $- 5$ to the power of $2$ .

$x = \frac{5 \pm \sqrt{25 - 4 \cdot 4 \cdot 3}}{2 \cdot 4}$

Step 6.1.2

Multiply $- 4 \cdot 4 \cdot 3$ .

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Step 6.1.2.1

Multiply $- 4$ by $4$ .

$x = \frac{5 \pm \sqrt{25 - 16 \cdot 3}}{2 \cdot 4}$

Step 6.1.2.2

Multiply $- 16$ by $3$ .

$x = \frac{5 \pm \sqrt{25 - 48}}{2 \cdot 4}$

Step 6.1.3

Subtract $48$ from $25$ .

$x = \frac{5 \pm \sqrt{- 23}}{2 \cdot 4}$

Step 6.1.4

Rewrite $- 23$ as $- 1 (23)$ .

$x = \frac{5 \pm \sqrt{- 1 \cdot 23}}{2 \cdot 4}$

Step 6.1.5

Rewrite $\sqrt{- 1 (23)}$ as $\sqrt{- 1} \cdot \sqrt{23}$ .

$x = \frac{5 \pm \sqrt{- 1} \cdot \sqrt{23}}{2 \cdot 4}$

Step 6.1.6

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

$x = \frac{5 \pm i \sqrt{23}}{2 \cdot 4}$

Step 6.2

Multiply $2$ by $4$ .

$x = \frac{5 \pm i \sqrt{23}}{8}$

Step 6.3

Change the $\pm$ to $+$ .

$x = \frac{5 + i \sqrt{23}}{8}$

Step 7

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 7.1

Simplify the numerator.

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Step 7.1.1

Raise $- 5$ to the power of $2$ .

$x = \frac{5 \pm \sqrt{25 - 4 \cdot 4 \cdot 3}}{2 \cdot 4}$

Step 7.1.2

Multiply $- 4 \cdot 4 \cdot 3$ .

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Step 7.1.2.1

Multiply $- 4$ by $4$ .

$x = \frac{5 \pm \sqrt{25 - 16 \cdot 3}}{2 \cdot 4}$

Step 7.1.2.2

Multiply $- 16$ by $3$ .

$x = \frac{5 \pm \sqrt{25 - 48}}{2 \cdot 4}$

Step 7.1.3

Subtract $48$ from $25$ .

$x = \frac{5 \pm \sqrt{- 23}}{2 \cdot 4}$

Step 7.1.4

Rewrite $- 23$ as $- 1 (23)$ .

$x = \frac{5 \pm \sqrt{- 1 \cdot 23}}{2 \cdot 4}$

Step 7.1.5

Rewrite $\sqrt{- 1 (23)}$ as $\sqrt{- 1} \cdot \sqrt{23}$ .

$x = \frac{5 \pm \sqrt{- 1} \cdot \sqrt{23}}{2 \cdot 4}$

Step 7.1.6

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

$x = \frac{5 \pm i \sqrt{23}}{2 \cdot 4}$

Step 7.2

Multiply $2$ by $4$ .

$x = \frac{5 \pm i \sqrt{23}}{8}$

Step 7.3

Change the $\pm$ to $-$ .

$x = \frac{5 - i \sqrt{23}}{8}$

Step 8

The final answer is the combination of both solutions.

$x = \frac{5 + i \sqrt{23}}{8}, \frac{5 - i \sqrt{23}}{8}$