Find the Roots (Zeros) 4x^2+29=-28x

$4 x^{2} + 29 = - 28 x$

Add $28 x$ to both sides of the equation.

$4 x^{2} + 29 + 28 x = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 4$ , $b = 28$ , and $c = 29$ into the quadratic formula and solve for $x$ .

$\frac{- 28 \pm \sqrt{28^{2} - 4 \cdot (4 \cdot 29)}}{2 \cdot 4}$

Simplify.

Simplify the numerator.

Raise $28$ to the power of $2$ .

$x = \frac{- 28 \pm \sqrt{784 - 4 \cdot 4 \cdot 29}}{2 \cdot 4}$

Multiply $- 4$ by $4$ .

$x = \frac{- 28 \pm \sqrt{784 - 16 \cdot 29}}{2 \cdot 4}$

Multiply $- 16$ by $29$ .

$x = \frac{- 28 \pm \sqrt{784 - 464}}{2 \cdot 4}$

Subtract $464$ from $784$ .

$x = \frac{- 28 \pm \sqrt{320}}{2 \cdot 4}$

Rewrite $320$ as $8^{2} \cdot 5$ .

Factor $64$ out of $320$ .

$x = \frac{- 28 \pm \sqrt{64 (5)}}{2 \cdot 4}$

Rewrite $64$ as $8^{2}$ .

$x = \frac{- 28 \pm \sqrt{8^{2} \cdot 5}}{2 \cdot 4}$

Pull terms out from under the radical.

$x = \frac{- 28 \pm 8 \sqrt{5}}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{- 28 \pm 8 \sqrt{5}}{8}$

Simplify $\frac{- 28 \pm 8 \sqrt{5}}{8}$ .

$x = \frac{- 7 \pm 2 \sqrt{5}}{2}$

The result can be shown in multiple forms.

Exact Form:

$x = \frac{- 7 \pm 2 \sqrt{5}}{2}$

Decimal Form:

$x = - 1.26393202 \dots, - 5.73606797 \dots$