Find All Complex Solutions 2x^2+14x-4=x^2+3x

$2 x^{2} + 14 x - 4 = x^{2} + 3 x$

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.

$2 x^{2} + 14 x - 4 - x^{2} = 3 x$

Subtract $3 x$ from both sides of the equation.

$2 x^{2} + 14 x - 4 - x^{2} - 3 x = 0$

Simplify $2 x^{2} + 14 x - 4 - x^{2} - 3 x$ .

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

$x^{2} + 14 x - 4 - 3 x = 0$

Subtract $3 x$ from $14 x$ .

$x^{2} + 11 x - 4 = 0$

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 $a = 1$ , $b = 11$ , and $c = - 4$ into the quadratic formula and solve for $x$ .

$\frac{- 11 \pm \sqrt{11^{2} - 4 \cdot (1 \cdot - 4)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $11$ to the power of $2$ .

$x = \frac{- 11 \pm \sqrt{121 - 4 \cdot 1 \cdot - 4}}{2 \cdot 1}$

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

Multiply $- 4$ by $1$ .

$x = \frac{- 11 \pm \sqrt{121 - 4 \cdot - 4}}{2 \cdot 1}$

Multiply $- 4$ by $- 4$ .

$x = \frac{- 11 \pm \sqrt{121 + 16}}{2 \cdot 1}$

Add $121$ and $16$ .

$x = \frac{- 11 \pm \sqrt{137}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 11 \pm \sqrt{137}}{2}$

Step 5

The final answer is the combination of both solutions.

$x = - \frac{11 - \sqrt{137}}{2}, - \frac{11 + \sqrt{137}}{2}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{11 - \sqrt{137}}{2}, - \frac{11 + \sqrt{137}}{2}$

Decimal Form:

$x = 0.35234995 \dots, - 11.35234995 \dots$