Solve Using the Quadratic Formula 2x^2+5x-10=x^2+4

$2 x^{2} + 5 x - 10 = x^{2} + 4$

Step 1

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

Move all the expressions to the left side of the equation.

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

$2 x^{2} + 5 x - 10 - x^{2} = 4$

Subtract $4$ from both sides of the equation.

$2 x^{2} + 5 x - 10 - x^{2} - 4 = 0$

Simplify $2 x^{2} + 5 x - 10 - x^{2} - 4$ .

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

$x^{2} + 5 x - 10 - 4 = 0$

Subtract $4$ from $- 10$ .

$x^{2} + 5 x - 14 = 0$

Step 2

Use the quadratic formula to find the solutions.

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

Step 3

Substitute the values $a = 1$ , $b = 5$ , and $c = - 14$ into the quadratic formula and solve for $x$ .

$\frac{- 5 \pm \sqrt{5^{2} - 4 \cdot (1 \cdot - 14)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $5$ to the power of $2$ .

$x = \frac{- 5 \pm \sqrt{25 - 4 \cdot 1 \cdot - 14}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 14$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 5 \pm \sqrt{25 - 4 \cdot - 14}}{2 \cdot 1}$

Multiply $- 4$ by $- 14$ .

$x = \frac{- 5 \pm \sqrt{25 + 56}}{2 \cdot 1}$

Add $25$ and $56$ .

$x = \frac{- 5 \pm \sqrt{81}}{2 \cdot 1}$

Rewrite $81$ as $9^{2}$ .

$x = \frac{- 5 \pm \sqrt{9^{2}}}{2 \cdot 1}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{- 5 \pm 9}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 5 \pm 9}{2}$

Step 5

The final answer is the combination of both solutions.

$x = 2, - 7$