Solve for x 2x^2+3x-7=x^2+5x+39

$2 x^{2} + 3 x - 7 = x^{2} + 5 x + 39$

Step 1

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

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

$2 x^{2} + 3 x - 7 - x^{2} = 5 x + 39$

Subtract $5 x$ from both sides of the equation.

$2 x^{2} + 3 x - 7 - x^{2} - 5 x = 39$

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

$x^{2} + 3 x - 7 - 5 x = 39$

Subtract $5 x$ from $3 x$ .

$x^{2} - 2 x - 7 = 39$

Step 2

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

Subtract $39$ from both sides of the equation.

$x^{2} - 2 x - 7 - 39 = 0$

Subtract $39$ from $- 7$ .

$x^{2} - 2 x - 46 = 0$

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 $a = 1$ , $b = - 2$ , and $c = - 46$ into the quadratic formula and solve for $x$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (1 \cdot - 46)}}{2 \cdot 1}$

Step 5

Simplify.

Simplify the numerator.

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

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 46}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 46$ .

Multiply $- 4$ by $1$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot - 46}}{2 \cdot 1}$

Multiply $- 4$ by $- 46$ .

$x = \frac{2 \pm \sqrt{4 + 184}}{2 \cdot 1}$

Add $4$ and $184$ .

$x = \frac{2 \pm \sqrt{188}}{2 \cdot 1}$

Rewrite $188$ as $2^{2} \cdot 47$ .

Factor $4$ out of $188$ .

$x = \frac{2 \pm \sqrt{4 (47)}}{2 \cdot 1}$

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

$x = \frac{2 \pm \sqrt{2^{2} \cdot 47}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{2 \pm 2 \sqrt{47}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{2 \pm 2 \sqrt{47}}{2}$

Simplify $\frac{2 \pm 2 \sqrt{47}}{2}$ .

$x = 1 \pm \sqrt{47}$

Step 6

The final answer is the combination of both solutions.

$x = 1 + \sqrt{47}, 1 - \sqrt{47}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$x = 1 + \sqrt{47}, 1 - \sqrt{47}$

Decimal Form:

$x = 7.85565460 \dots, - 5.85565460 \dots$

Solve for x 2x^2+3x-7=x^2+5x+39

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