Solve Using the Quadratic Formula 3x^2=10x+48

$3 x^{2} = 10 x + 48$

Step 1

Move all the expressions to the left side of the equation.

Subtract $10 x$ from both sides of the equation.

$3 x^{2} - 10 x = 48$

Subtract $48$ from both sides of the equation.

$3 x^{2} - 10 x - 48 = 0$

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 $a = 3$ , $b = - 10$ , and $c = - 48$ into the quadratic formula and solve for $x$ .

$\frac{10 \pm \sqrt{{(- 10)}^{2} - 4 \cdot (3 \cdot - 48)}}{2 \cdot 3}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 10$ to the power of $2$ .

$x = \frac{10 \pm \sqrt{100 - 4 \cdot 3 \cdot - 48}}{2 \cdot 3}$

Multiply $- 4 \cdot 3 \cdot - 48$ .

Multiply $- 4$ by $3$ .

$x = \frac{10 \pm \sqrt{100 - 12 \cdot - 48}}{2 \cdot 3}$

Multiply $- 12$ by $- 48$ .

$x = \frac{10 \pm \sqrt{100 + 576}}{2 \cdot 3}$

Add $100$ and $576$ .

$x = \frac{10 \pm \sqrt{676}}{2 \cdot 3}$

Rewrite $676$ as $26^{2}$ .

$x = \frac{10 \pm \sqrt{26^{2}}}{2 \cdot 3}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{10 \pm 26}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{10 \pm 26}{6}$

Simplify $\frac{10 \pm 26}{6}$ .

$x = \frac{5 \pm 13}{3}$

Step 5

The final answer is the combination of both solutions.

$x = 6, - \frac{8}{3}$