Find the Roots (Zeros) -3x+19=-x^2+1

$- 3 x + 19 = - x^{2} + 1$

Step 1

Add $x^{2}$ to both sides of the equation.

$- 3 x + 19 + x^{2} = 1$

Step 2

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

Subtract $1$ from both sides of the equation.

$- 3 x + 19 + x^{2} - 1 = 0$

Subtract $1$ from $19$ .

$- 3 x + x^{2} + 18 = 0$

Step 3

Use the quadratic formula to find the solutions.

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

Step 4

Substitute the values $a = 1$ , $b = - 3$ , and $c = 18$ into the quadratic formula and solve for $x$ .

$\frac{3 \pm \sqrt{{(- 3)}^{2} - 4 \cdot (1 \cdot 18)}}{2 \cdot 1}$

Step 5

Simplify.

Simplify the numerator.

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

$x = \frac{3 \pm \sqrt{9 - 4 \cdot 1 \cdot 18}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 18$ .

Multiply $- 4$ by $1$ .

$x = \frac{3 \pm \sqrt{9 - 4 \cdot 18}}{2 \cdot 1}$

Multiply $- 4$ by $18$ .

$x = \frac{3 \pm \sqrt{9 - 72}}{2 \cdot 1}$

Subtract $72$ from $9$ .

$x = \frac{3 \pm \sqrt{- 63}}{2 \cdot 1}$

Rewrite $- 63$ as $- 1 (63)$ .

$x = \frac{3 \pm \sqrt{- 1 \cdot 63}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (63)}$ as $\sqrt{- 1} \cdot \sqrt{63}$ .

$x = \frac{3 \pm \sqrt{- 1} \cdot \sqrt{63}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{3 \pm i \cdot \sqrt{63}}{2 \cdot 1}$

Rewrite $63$ as $3^{2} \cdot 7$ .

Factor $9$ out of $63$ .

$x = \frac{3 \pm i \cdot \sqrt{9 (7)}}{2 \cdot 1}$

Rewrite $9$ as $3^{2}$ .

$x = \frac{3 \pm i \cdot \sqrt{3^{2} \cdot 7}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{3 \pm i \cdot (3 \sqrt{7})}{2 \cdot 1}$

Move $3$ to the left of $i$ .

$x = \frac{3 \pm 3 i \sqrt{7}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{3 \pm 3 i \sqrt{7}}{2}$

Step 6

Find the Roots (Zeros) -3x+19=-x^2+1

Related Questions

Support LCMGCF.com by disabling your adblocker.