Precalculus Examples

$10 x^{2} = 13 x + 3$

Step 1

Move all the expressions to the left side of the equation.

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Step 1.1

Subtract $13 x$ from both sides of the equation.

$10 x^{2} - 13 x = 3$

Step 1.2

Subtract $3$ from both sides of the equation.

$10 x^{2} - 13 x - 3 = 0$

Step 2

Use the quadratic formula to find the solutions.

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

Step 3

Substitute the values $a = 10$ , $b = - 13$ , and $c = - 3$ into the quadratic formula and solve for $x$ .

$\frac{13 \pm \sqrt{{(- 13)}^{2} - 4 \cdot (10 \cdot - 3)}}{2 \cdot 10}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Raise $- 13$ to the power of $2$ .

$x = \frac{13 \pm \sqrt{169 - 4 \cdot 10 \cdot - 3}}{2 \cdot 10}$

Step 4.1.2

Multiply $- 4 \cdot 10 \cdot - 3$ .

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Step 4.1.2.1

Multiply $- 4$ by $10$ .

$x = \frac{13 \pm \sqrt{169 - 40 \cdot - 3}}{2 \cdot 10}$

Step 4.1.2.2

Multiply $- 40$ by $- 3$ .

$x = \frac{13 \pm \sqrt{169 + 120}}{2 \cdot 10}$

Step 4.1.3

Add $169$ and $120$ .

$x = \frac{13 \pm \sqrt{289}}{2 \cdot 10}$

Step 4.1.4

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

$x = \frac{13 \pm \sqrt{17^{2}}}{2 \cdot 10}$

Step 4.1.5

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

$x = \frac{13 \pm 17}{2 \cdot 10}$

Step 4.2

Multiply $2$ by $10$ .

$x = \frac{13 \pm 17}{20}$

Step 5

The final answer is the combination of both solutions.

$x = \frac{3}{2}, - \frac{1}{5}$