Solve Using the Quadratic Formula x^2-3x+1=6

$x^{2} - 3 x + 1 = 6$

Step 1

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

Subtract $6$ from both sides of the equation.

$x^{2} - 3 x + 1 - 6 = 0$

Subtract $6$ from $1$ .

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

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 $a = 1$ , $b = - 3$ , and $c = - 5$ into the quadratic formula and solve for $x$ .

$\frac{3 \pm \sqrt{{(- 3)}^{2} - 4 \cdot (1 \cdot - 5)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 3$ to the power of $2$ .

$x = \frac{3 \pm \sqrt{9 - 4 \cdot 1 \cdot - 5}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 5$ .

Multiply $- 4$ by $1$ .

$x = \frac{3 \pm \sqrt{9 - 4 \cdot - 5}}{2 \cdot 1}$

Multiply $- 4$ by $- 5$ .

$x = \frac{3 \pm \sqrt{9 + 20}}{2 \cdot 1}$

Add $9$ and $20$ .

$x = \frac{3 \pm \sqrt{29}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{3 \pm \sqrt{29}}{2}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = \frac{3 \pm \sqrt{29}}{2}$

Decimal Form:

$x = 4.19258240 \dots, - 1.19258240 \dots$