Algebra Examples

$10 x^{2} = 6 x + 3$

Step 1

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

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Step 1.1

Subtract $6 x$ from both sides of the equation.

$10 x^{2} - 6 x = 3$

Step 1.2

Subtract $3$ from both sides of the equation.

$10 x^{2} - 6 x - 3 = 0$

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 $a = 10$ , $b = - 6$ , and $c = - 3$ into the quadratic formula and solve for $x$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (10 \cdot - 3)}}{2 \cdot 10}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Raise $- 6$ to the power of $2$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 10 \cdot - 3}}{2 \cdot 10}$

Step 4.1.2

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

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Step 4.1.2.1

Multiply $- 4$ by $10$ .

$x = \frac{6 \pm \sqrt{36 - 40 \cdot - 3}}{2 \cdot 10}$

Step 4.1.2.2

Multiply $- 40$ by $- 3$ .

$x = \frac{6 \pm \sqrt{36 + 120}}{2 \cdot 10}$

Step 4.1.3

Add $36$ and $120$ .

$x = \frac{6 \pm \sqrt{156}}{2 \cdot 10}$

Step 4.1.4

Rewrite $156$ as $2^{2} \cdot 39$ .

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Step 4.1.4.1

Factor $4$ out of $156$ .

$x = \frac{6 \pm \sqrt{4 (39)}}{2 \cdot 10}$

Step 4.1.4.2

Rewrite $4$ as $2^{2}$ .

$x = \frac{6 \pm \sqrt{2^{2} \cdot 39}}{2 \cdot 10}$

Step 4.1.5

Pull terms out from under the radical.

$x = \frac{6 \pm 2 \sqrt{39}}{2 \cdot 10}$

Step 4.2

Multiply $2$ by $10$ .

$x = \frac{6 \pm 2 \sqrt{39}}{20}$

Step 4.3

Simplify $\frac{6 \pm 2 \sqrt{39}}{20}$ .

$x = \frac{3 \pm \sqrt{39}}{10}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = \frac{3 \pm \sqrt{39}}{10}$

Decimal Form:

$x = 0.92449979 \dots, - 0.32449979 \dots$