Find the Roots (Zeros) 15x^2-3x=5x^2-1

$15 x^{2} - 3 x = 5 x^{2} - 1$

Step 1

Move all terms containing $x$ to the left side of the equation.

Subtract $5 x^{2}$ from both sides of the equation.

$15 x^{2} - 3 x - 5 x^{2} = - 1$

Subtract $5 x^{2}$ from $15 x^{2}$ .

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

Step 2

Add $1$ to both sides of the equation.

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

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 $a = 10$ , $b = - 3$ , and $c = 1$ into the quadratic formula and solve for $x$ .

$\frac{3 \pm \sqrt{{(- 3)}^{2} - 4 \cdot (10 \cdot 1)}}{2 \cdot 10}$

Step 5

Simplify.

Simplify the numerator.

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

$x = \frac{3 \pm \sqrt{9 - 4 \cdot 10 \cdot 1}}{2 \cdot 10}$

Multiply $- 4 \cdot 10 \cdot 1$ .

Multiply $- 4$ by $10$ .

$x = \frac{3 \pm \sqrt{9 - 40 \cdot 1}}{2 \cdot 10}$

Multiply $- 40$ by $1$ .

$x = \frac{3 \pm \sqrt{9 - 40}}{2 \cdot 10}$

Subtract $40$ from $9$ .

$x = \frac{3 \pm \sqrt{- 31}}{2 \cdot 10}$

Rewrite $- 31$ as $- 1 (31)$ .

$x = \frac{3 \pm \sqrt{- 1 \cdot 31}}{2 \cdot 10}$

Rewrite $\sqrt{- 1 (31)}$ as $\sqrt{- 1} \cdot \sqrt{31}$ .

$x = \frac{3 \pm \sqrt{- 1} \cdot \sqrt{31}}{2 \cdot 10}$

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

$x = \frac{3 \pm i \sqrt{31}}{2 \cdot 10}$

Multiply $2$ by $10$ .

$x = \frac{3 \pm i \sqrt{31}}{20}$

Step 6