Find the Roots (Zeros) 3x^2-16x+15=3

$3 x^{2} - 16 x + 15 = 3$

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

Subtract $3$ from both sides of the equation.

$3 x^{2} - 16 x + 15 - 3 = 0$

Subtract $3$ from $15$ .

$3 x^{2} - 16 x + 12 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 3$ , $b = - 16$ , and $c = 12$ into the quadratic formula and solve for $x$ .

$\frac{16 \pm \sqrt{{(- 16)}^{2} - 4 \cdot (3 \cdot 12)}}{2 \cdot 3}$

Simplify.

Simplify the numerator.

Raise $- 16$ to the power of $2$ .

$x = \frac{16 \pm \sqrt{256 - 4 \cdot 3 \cdot 12}}{2 \cdot 3}$

Multiply $- 4$ by $3$ .

$x = \frac{16 \pm \sqrt{256 - 12 \cdot 12}}{2 \cdot 3}$

Multiply $- 12$ by $12$ .

$x = \frac{16 \pm \sqrt{256 - 144}}{2 \cdot 3}$

Subtract $144$ from $256$ .

$x = \frac{16 \pm \sqrt{112}}{2 \cdot 3}$

Rewrite $112$ as $4^{2} \cdot 7$ .

Factor $16$ out of $112$ .

$x = \frac{16 \pm \sqrt{16 (7)}}{2 \cdot 3}$

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

$x = \frac{16 \pm \sqrt{4^{2} \cdot 7}}{2 \cdot 3}$

Pull terms out from under the radical.

$x = \frac{16 \pm 4 \sqrt{7}}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{16 \pm 4 \sqrt{7}}{6}$

Simplify $\frac{16 \pm 4 \sqrt{7}}{6}$ .

$x = \frac{8 \pm 2 \sqrt{7}}{3}$

The result can be shown in multiple forms.

Exact Form:

$x = \frac{8 \pm 2 \sqrt{7}}{3}$

Decimal Form:

$x = 4.43050087 \dots, 0.90283245 \dots$