Find the Roots (Zeros) 10x^2-6x+4=-3x

$10 x^{2} - 6 x + 4 = - 3 x$

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

Add $3 x$ to both sides of the equation.

$10 x^{2} - 6 x + 4 + 3 x = 0$

Add $- 6 x$ and $3 x$ .

$10 x^{2} - 3 x + 4 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 10$ , $b = - 3$ , and $c = 4$ into the quadratic formula and solve for $x$ .

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

Simplify.

Simplify the numerator.

Raise $- 3$ to the power of $2$ .

$x = \frac{3 \pm \sqrt{9 - 4 \cdot 10 \cdot 4}}{2 \cdot 10}$

Multiply $- 4$ by $10$ .

$x = \frac{3 \pm \sqrt{9 - 40 \cdot 4}}{2 \cdot 10}$

Multiply $- 40$ by $4$ .

$x = \frac{3 \pm \sqrt{9 - 160}}{2 \cdot 10}$

Subtract $160$ from $9$ .

$x = \frac{3 \pm \sqrt{- 151}}{2 \cdot 10}$

Rewrite $- 151$ as $- 1 (151)$ .

$x = \frac{3 \pm \sqrt{- 1 \cdot 151}}{2 \cdot 10}$

Rewrite $\sqrt{- 1 (151)}$ as $\sqrt{- 1} \cdot \sqrt{151}$ .

$x = \frac{3 \pm \sqrt{- 1} \cdot \sqrt{151}}{2 \cdot 10}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{3 \pm i \sqrt{151}}{2 \cdot 10}$

Multiply $2$ by $10$ .

$x = \frac{3 \pm i \sqrt{151}}{20}$