Find the Roots (Zeros) 7x^2-9x+12=5x^2-6

$7 x^{2} - 9 x + 12 = 5 x^{2} - 6$

Step 1

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

Subtract $5 x^{2}$ from both sides of the equation.

$7 x^{2} - 9 x + 12 - 5 x^{2} = - 6$

Subtract $5 x^{2}$ from $7 x^{2}$ .

$2 x^{2} - 9 x + 12 = - 6$

Step 2

Move all terms to the left side of the equation and simplify.

Add $6$ to both sides of the equation.

$2 x^{2} - 9 x + 12 + 6 = 0$

Add $12$ and $6$ .

$2 x^{2} - 9 x + 18 = 0$

Step 3

Use the quadratic formula to find the solutions.

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

Step 4

Substitute the values $a = 2$ , $b = - 9$ , and $c = 18$ into the quadratic formula and solve for $x$ .

$\frac{9 \pm \sqrt{{(- 9)}^{2} - 4 \cdot (2 \cdot 18)}}{2 \cdot 2}$

Step 5

Simplify.

Simplify the numerator.

Raise $- 9$ to the power of $2$ .

$x = \frac{9 \pm \sqrt{81 - 4 \cdot 2 \cdot 18}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot 18$ .

Multiply $- 4$ by $2$ .

$x = \frac{9 \pm \sqrt{81 - 8 \cdot 18}}{2 \cdot 2}$

Multiply $- 8$ by $18$ .

$x = \frac{9 \pm \sqrt{81 - 144}}{2 \cdot 2}$

Subtract $144$ from $81$ .

$x = \frac{9 \pm \sqrt{- 63}}{2 \cdot 2}$

Rewrite $- 63$ as $- 1 (63)$ .

$x = \frac{9 \pm \sqrt{- 1 \cdot 63}}{2 \cdot 2}$

Rewrite $\sqrt{- 1 (63)}$ as $\sqrt{- 1} \cdot \sqrt{63}$ .

$x = \frac{9 \pm \sqrt{- 1} \cdot \sqrt{63}}{2 \cdot 2}$

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

$x = \frac{9 \pm i \cdot \sqrt{63}}{2 \cdot 2}$

Rewrite $63$ as $3^{2} \cdot 7$ .

Factor $9$ out of $63$ .

$x = \frac{9 \pm i \cdot \sqrt{9 (7)}}{2 \cdot 2}$

Rewrite $9$ as $3^{2}$ .

$x = \frac{9 \pm i \cdot \sqrt{3^{2} \cdot 7}}{2 \cdot 2}$

Pull terms out from under the radical.

$x = \frac{9 \pm i \cdot (3 \sqrt{7})}{2 \cdot 2}$

Move $3$ to the left of $i$ .

$x = \frac{9 \pm 3 i \sqrt{7}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{9 \pm 3 i \sqrt{7}}{4}$

Step 6

Find the Roots (Zeros) 7x^2-9x+12=5x^2-6

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