Find the Roots (Zeros) 5x^2-6x-1=-3x^2

$5 x^{2} - 6 x - 1 = - 3 x^{2}$

Move all terms containing $x$ to the left side of the equation.

Add $3 x^{2}$ to both sides of the equation.

$5 x^{2} - 6 x - 1 + 3 x^{2} = 0$

Add $5 x^{2}$ and $3 x^{2}$ .

$8 x^{2} - 6 x - 1 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 8$ , $b = - 6$ , and $c = - 1$ into the quadratic formula and solve for $x$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (8 \cdot - 1)}}{2 \cdot 8}$

Simplify.

Simplify the numerator.

Raise $- 6$ to the power of $2$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 8 \cdot - 1}}{2 \cdot 8}$

Multiply $- 4$ by $8$ .

$x = \frac{6 \pm \sqrt{36 - 32 \cdot - 1}}{2 \cdot 8}$

Multiply $- 32$ by $- 1$ .

$x = \frac{6 \pm \sqrt{36 + 32}}{2 \cdot 8}$

Add $36$ and $32$ .

$x = \frac{6 \pm \sqrt{68}}{2 \cdot 8}$

Rewrite $68$ as $2^{2} \cdot 17$ .

Factor $4$ out of $68$ .

$x = \frac{6 \pm \sqrt{4 (17)}}{2 \cdot 8}$

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

$x = \frac{6 \pm \sqrt{2^{2} \cdot 17}}{2 \cdot 8}$

Pull terms out from under the radical.

$x = \frac{6 \pm 2 \sqrt{17}}{2 \cdot 8}$

Multiply $2$ by $8$ .

$x = \frac{6 \pm 2 \sqrt{17}}{16}$

Simplify $\frac{6 \pm 2 \sqrt{17}}{16}$ .

$x = \frac{3 \pm \sqrt{17}}{8}$

The result can be shown in multiple forms.

Exact Form:

$x = \frac{3 \pm \sqrt{17}}{8}$

Decimal Form:

$x = 0.89038820 \dots, - 0.14038820 \dots$