Find the Roots (Zeros) 9x^2=4+7x

$9 x^{2} = 4 + 7 x$

Subtract $7 x$ from both sides of the equation.

$9 x^{2} - 7 x = 4$

Subtract $4$ from both sides of the equation.

$9 x^{2} - 7 x - 4 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 9$ , $b = - 7$ , and $c = - 4$ into the quadratic formula and solve for $x$ .

$\frac{7 \pm \sqrt{{(- 7)}^{2} - 4 \cdot (9 \cdot - 4)}}{2 \cdot 9}$

Simplify.

Simplify the numerator.

Raise $- 7$ to the power of $2$ .

$x = \frac{7 \pm \sqrt{49 - 4 \cdot 9 \cdot - 4}}{2 \cdot 9}$

Multiply $- 4$ by $9$ .

$x = \frac{7 \pm \sqrt{49 - 36 \cdot - 4}}{2 \cdot 9}$

Multiply $- 36$ by $- 4$ .

$x = \frac{7 \pm \sqrt{49 + 144}}{2 \cdot 9}$

Add $49$ and $144$ .

$x = \frac{7 \pm \sqrt{193}}{2 \cdot 9}$

Multiply $2$ by $9$ .

$x = \frac{7 \pm \sqrt{193}}{18}$

The result can be shown in multiple forms.

Exact Form:

$x = \frac{7 \pm \sqrt{193}}{18}$

Decimal Form:

$x = 1.16069133 \dots, - 0.38291355 \dots$