Solve Using the Quadratic Formula 15a^2-3a=3-7a

$15 a^{2} - 3 a = 3 - 7 a$

Step 1

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

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

Subtract $3$ from both sides of the equation.

$15 a^{2} - 3 a - 3 = - 7 a$

Add $7 a$ to both sides of the equation.

$15 a^{2} - 3 a - 3 + 7 a = 0$

Add $- 3 a$ and $7 a$ .

$15 a^{2} + 4 a - 3 = 0$

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 $a = 15$ , $b = 4$ , and $c = - 3$ into the quadratic formula and solve for $a$ .

$\frac{- 4 \pm \sqrt{4^{2} - 4 \cdot (15 \cdot - 3)}}{2 \cdot 15}$

Step 4

Simplify.

Simplify the numerator.

Raise $4$ to the power of $2$ .

$a = \frac{- 4 \pm \sqrt{16 - 4 \cdot 15 \cdot - 3}}{2 \cdot 15}$

Multiply $- 4 \cdot 15 \cdot - 3$ .

Multiply $- 4$ by $15$ .

$a = \frac{- 4 \pm \sqrt{16 - 60 \cdot - 3}}{2 \cdot 15}$

Multiply $- 60$ by $- 3$ .

$a = \frac{- 4 \pm \sqrt{16 + 180}}{2 \cdot 15}$

Add $16$ and $180$ .

$a = \frac{- 4 \pm \sqrt{196}}{2 \cdot 15}$

Rewrite $196$ as $14^{2}$ .

$a = \frac{- 4 \pm \sqrt{14^{2}}}{2 \cdot 15}$

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

$a = \frac{- 4 \pm 14}{2 \cdot 15}$

Multiply $2$ by $15$ .

$a = \frac{- 4 \pm 14}{30}$

Simplify $\frac{- 4 \pm 14}{30}$ .

$a = \frac{- 2 \pm 7}{15}$

Step 5

The final answer is the combination of both solutions.

$a = \frac{1}{3}, - \frac{3}{5}$