Solve the System of Equations y=x^2-3 y=-2x+4

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

Eliminate the equal sides of each equation and combine.

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

Solve $x^{2} - 3 = - 2 x + 4$ for $x$ .

Add $2 x$ to both sides of the equation.

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

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

Subtract $4$ from both sides of the equation.

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

Subtract $4$ from $- 3$ .

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

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = 2$ , and $c = - 7$ into the quadratic formula and solve for $x$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (1 \cdot - 7)}}{2 \cdot 1} + y = - 2 x + 4$

Simplify.

Simplify the numerator.

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

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 7}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 7}}{2 \cdot 1}$

Multiply $- 4$ by $- 7$ .

$x = \frac{- 2 \pm \sqrt{4 + 28}}{2 \cdot 1}$

Add $4$ and $28$ .

$x = \frac{- 2 \pm \sqrt{32}}{2 \cdot 1}$

Rewrite $32$ as $4^{2} \cdot 2$ .

Factor $16$ out of $32$ .

$x = \frac{- 2 \pm \sqrt{16 (2)}}{2 \cdot 1}$

Rewrite $16$ as $4^{2}$ .

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

Pull terms out from under the radical.

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

Multiply $2$ by $1$ .

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

Simplify $\frac{- 2 \pm 4 \sqrt{2}}{2}$ .

$x = - 1 \pm 2 \sqrt{2}$

Simplify the expression to solve for the $+$ portion of the $\pm$ .

Simplify the numerator.

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

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 7}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 7}}{2 \cdot 1}$

Multiply $- 4$ by $- 7$ .

$x = \frac{- 2 \pm \sqrt{4 + 28}}{2 \cdot 1}$

Add $4$ and $28$ .

$x = \frac{- 2 \pm \sqrt{32}}{2 \cdot 1}$

Rewrite $32$ as $4^{2} \cdot 2$ .

Factor $16$ out of $32$ .

$x = \frac{- 2 \pm \sqrt{16 (2)}}{2 \cdot 1}$

Rewrite $16$ as $4^{2}$ .

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

Pull terms out from under the radical.

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

Multiply $2$ by $1$ .

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

Simplify $\frac{- 2 \pm 4 \sqrt{2}}{2}$ .

$x = - 1 \pm 2 \sqrt{2}$

Change the $\pm$ to $+$ .

$x = - 1 + 2 \sqrt{2}$

Simplify the expression to solve for the $-$ portion of the $\pm$ .

Simplify the numerator.

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

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 7}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 7}}{2 \cdot 1}$

Multiply $- 4$ by $- 7$ .

$x = \frac{- 2 \pm \sqrt{4 + 28}}{2 \cdot 1}$

Add $4$ and $28$ .

$x = \frac{- 2 \pm \sqrt{32}}{2 \cdot 1}$

Rewrite $32$ as $4^{2} \cdot 2$ .

Factor $16$ out of $32$ .

$x = \frac{- 2 \pm \sqrt{16 (2)}}{2 \cdot 1}$

Rewrite $16$ as $4^{2}$ .

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

Pull terms out from under the radical.

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

Multiply $2$ by $1$ .

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

Simplify $\frac{- 2 \pm 4 \sqrt{2}}{2}$ .

$x = - 1 \pm 2 \sqrt{2}$

Change the $\pm$ to $-$ .

$x = - 1 - 2 \sqrt{2}$

The final answer is the combination of both solutions.

$x = - 1 + 2 \sqrt{2}, - 1 - 2 \sqrt{2}$

Evaluate $y$ when $x = - 1 + 2 \sqrt{2}$ .

Substitute $- 1 + 2 \sqrt{2}$ for $x$ .

$y = - 2 (- 1 + 2 \sqrt{2}) + 4$

Simplify $- 2 (- 1 + 2 \sqrt{2}) + 4$ .

Simplify each term.

Apply the distributive property.

$y = - 2 \cdot - 1 - 2 (2 \sqrt{2}) + 4$

Multiply $- 2$ by $- 1$ .

$y = 2 - 2 (2 \sqrt{2}) + 4$

Multiply $2$ by $- 2$ .

$y = 2 - 4 \sqrt{2} + 4$

Add $2$ and $4$ .

$y = 6 - 4 \sqrt{2}$

Evaluate $y$ when $x = - 1 - 2 \sqrt{2}$ .

Substitute $- 1 - 2 \sqrt{2}$ for $x$ .

$y = - 2 (- 1 - 2 \sqrt{2}) + 4$

Simplify $- 2 (- 1 - 2 \sqrt{2}) + 4$ .

Simplify each term.

Apply the distributive property.

$y = - 2 \cdot - 1 - 2 (- 2 \sqrt{2}) + 4$

Multiply $- 2$ by $- 1$ .

$y = 2 - 2 (- 2 \sqrt{2}) + 4$

Multiply $- 2$ by $- 2$ .

$y = 2 + 4 \sqrt{2} + 4$

Add $2$ and $4$ .

$y = 6 + 4 \sqrt{2}$

The solution to the system is the complete set of ordered pairs that are valid solutions.

$(- 1 + 2 \sqrt{2}, 6 - 4 \sqrt{2})$

$(- 1 - 2 \sqrt{2}, 6 + 4 \sqrt{2})$

The result can be shown in multiple forms.

Point Form:

$(- 1 + 2 \sqrt{2}, 6 - 4 \sqrt{2}), (- 1 - 2 \sqrt{2}, 6 + 4 \sqrt{2})$

Equation Form:

$x = - 1 + 2 \sqrt{2}, y = 6 - 4 \sqrt{2}$

$x = - 1 - 2 \sqrt{2}, y = 6 + 4 \sqrt{2}$

Solve the System of Equations y=x^2-3 y=-2x+4

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