Solve Using the Quadratic Formula 5x^2-2x+13=18x+1

$5 x^{2} - 2 x + 13 = 18 x + 1$

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 $18 x$ from both sides of the equation.

$5 x^{2} - 2 x + 13 - 18 x = 1$

Subtract $1$ from both sides of the equation.

$5 x^{2} - 2 x + 13 - 18 x - 1 = 0$

Simplify $5 x^{2} - 2 x + 13 - 18 x - 1$ .

Subtract $18 x$ from $- 2 x$ .

$5 x^{2} - 20 x + 13 - 1 = 0$

Subtract $1$ from $13$ .

$5 x^{2} - 20 x + 12 = 0$

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 $a = 5$ , $b = - 20$ , and $c = 12$ into the quadratic formula and solve for $x$ .

$\frac{20 \pm \sqrt{{(- 20)}^{2} - 4 \cdot (5 \cdot 12)}}{2 \cdot 5}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 20$ to the power of $2$ .

$x = \frac{20 \pm \sqrt{400 - 4 \cdot 5 \cdot 12}}{2 \cdot 5}$

Multiply $- 4 \cdot 5 \cdot 12$ .

Multiply $- 4$ by $5$ .

$x = \frac{20 \pm \sqrt{400 - 20 \cdot 12}}{2 \cdot 5}$

Multiply $- 20$ by $12$ .

$x = \frac{20 \pm \sqrt{400 - 240}}{2 \cdot 5}$

Subtract $240$ from $400$ .

$x = \frac{20 \pm \sqrt{160}}{2 \cdot 5}$

Rewrite $160$ as $4^{2} \cdot 10$ .

Factor $16$ out of $160$ .

$x = \frac{20 \pm \sqrt{16 (10)}}{2 \cdot 5}$

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

$x = \frac{20 \pm \sqrt{4^{2} \cdot 10}}{2 \cdot 5}$

Pull terms out from under the radical.

$x = \frac{20 \pm 4 \sqrt{10}}{2 \cdot 5}$

Multiply $2$ by $5$ .

$x = \frac{20 \pm 4 \sqrt{10}}{10}$

Simplify $\frac{20 \pm 4 \sqrt{10}}{10}$ .

$x = \frac{10 \pm 2 \sqrt{10}}{5}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = \frac{10 \pm 2 \sqrt{10}}{5}$

Decimal Form:

$x = 3.26491106 \dots, 0.73508893 \dots$

Solve Using the Quadratic Formula 5x^2-2x+13=18x+1

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