Solve Using the Quadratic Formula 3x^2-2x+5=10x+1

$3 x^{2} - 2 x + 5 = 10 x + 1$

Step 1

Move all terms to the left side of the equation and simplify.

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

Subtract $10 x$ from both sides of the equation.

$3 x^{2} - 2 x + 5 - 10 x = 1$

Subtract $1$ from both sides of the equation.

$3 x^{2} - 2 x + 5 - 10 x - 1 = 0$

Simplify $3 x^{2} - 2 x + 5 - 10 x - 1$ .

Subtract $10 x$ from $- 2 x$ .

$3 x^{2} - 12 x + 5 - 1 = 0$

Subtract $1$ from $5$ .

$3 x^{2} - 12 x + 4 = 0$

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 $a = 3$ , $b = - 12$ , and $c = 4$ into the quadratic formula and solve for $x$ .

$\frac{12 \pm \sqrt{{(- 12)}^{2} - 4 \cdot (3 \cdot 4)}}{2 \cdot 3}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 12$ to the power of $2$ .

$x = \frac{12 \pm \sqrt{144 - 4 \cdot 3 \cdot 4}}{2 \cdot 3}$

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

Multiply $- 4$ by $3$ .

$x = \frac{12 \pm \sqrt{144 - 12 \cdot 4}}{2 \cdot 3}$

Multiply $- 12$ by $4$ .

$x = \frac{12 \pm \sqrt{144 - 48}}{2 \cdot 3}$

Subtract $48$ from $144$ .

$x = \frac{12 \pm \sqrt{96}}{2 \cdot 3}$

Rewrite $96$ as $4^{2} \cdot 6$ .

Factor $16$ out of $96$ .

$x = \frac{12 \pm \sqrt{16 (6)}}{2 \cdot 3}$

Rewrite $16$ as $4^{2}$ .

$x = \frac{12 \pm \sqrt{4^{2} \cdot 6}}{2 \cdot 3}$

Pull terms out from under the radical.

$x = \frac{12 \pm 4 \sqrt{6}}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{12 \pm 4 \sqrt{6}}{6}$

Simplify $\frac{12 \pm 4 \sqrt{6}}{6}$ .

$x = \frac{6 \pm 2 \sqrt{6}}{3}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = \frac{6 \pm 2 \sqrt{6}}{3}$

Decimal Form:

$x = 3.63299316 \dots, 0.36700683 \dots$

Solve Using the Quadratic Formula 3x^2-2x+5=10x+1

Related Questions

Support LCMGCF.com by disabling your adblocker.