Solve Using the Quadratic Formula 4x^2+x=-7x-3

$4 x^{2} + x = - 7 x - 3$

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.

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

$4 x^{2} + x + 7 x = - 3$

Add $3$ to both sides of the equation.

$4 x^{2} + x + 7 x + 3 = 0$

Add $x$ and $7 x$ .

$4 x^{2} + 8 x + 3 = 0$

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 $a = 4$ , $b = 8$ , and $c = 3$ into the quadratic formula and solve for $x$ .

$\frac{- 8 \pm \sqrt{8^{2} - 4 \cdot (4 \cdot 3)}}{2 \cdot 4}$

Step 4

Simplify.

Simplify the numerator.

Raise $8$ to the power of $2$ .

$x = \frac{- 8 \pm \sqrt{64 - 4 \cdot 4 \cdot 3}}{2 \cdot 4}$

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

Multiply $- 4$ by $4$ .

$x = \frac{- 8 \pm \sqrt{64 - 16 \cdot 3}}{2 \cdot 4}$

Multiply $- 16$ by $3$ .

$x = \frac{- 8 \pm \sqrt{64 - 48}}{2 \cdot 4}$

Subtract $48$ from $64$ .

$x = \frac{- 8 \pm \sqrt{16}}{2 \cdot 4}$

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

$x = \frac{- 8 \pm \sqrt{4^{2}}}{2 \cdot 4}$

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

$x = \frac{- 8 \pm 4}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{- 8 \pm 4}{8}$

Simplify $\frac{- 8 \pm 4}{8}$ .

$x = \frac{- 2 \pm 1}{2}$

Step 5

The final answer is the combination of both solutions.

$x = - \frac{1}{2}, - \frac{3}{2}$