Solve C -2x^ 2 +x+7=-x^ 2 +5x+p=x^ 2 ) | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$- 2 x ^ { 2 } + x + 7 = - x ^ { 2 } + 5 x + ( - 1 )$$

Solve for x

$x=2\sqrt{3}-2\approx 1.464101615$
$x=-2\sqrt{3}-2\approx -5.464101615$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Add $x^{2}$ to both sides.

$$-2x^{2}+x+7+x^{2}=5x-1$$

Combine $-2x^{2}$ and $x^{2}$ to get $-x^{2}$.

$$-x^{2}+x+7=5x-1$$

Subtract $5x$ from both sides.

$$-x^{2}+x+7-5x=-1$$

Combine $x$ and $-5x$ to get $-4x$.

$$-x^{2}-4x+7=-1$$

Add $1$ to both sides.

$$-x^{2}-4x+7+1=0$$

Add $7$ and $1$ to get $8$.

$$-x^{2}-4x+8=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-1$ for $a$, $-4$ for $b$, and $8$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-1\right)\times 8}}{2\left(-1\right)}$$

Square $-4$.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-1\right)\times 8}}{2\left(-1\right)}$$

Multiply $-4$ times $-1$.

$$x=\frac{-\left(-4\right)±\sqrt{16+4\times 8}}{2\left(-1\right)}$$

Multiply $4$ times $8$.

$$x=\frac{-\left(-4\right)±\sqrt{16+32}}{2\left(-1\right)}$$

Add $16$ to $32$.

$$x=\frac{-\left(-4\right)±\sqrt{48}}{2\left(-1\right)}$$

Take the square root of $48$.

$$x=\frac{-\left(-4\right)±4\sqrt{3}}{2\left(-1\right)}$$

The opposite of $-4$ is $4$.

$$x=\frac{4±4\sqrt{3}}{2\left(-1\right)}$$

Multiply $2$ times $-1$.

$$x=\frac{4±4\sqrt{3}}{-2}$$

Now solve the equation $x=\frac{4±4\sqrt{3}}{-2}$ when $±$ is plus. Add $4$ to $4\sqrt{3}$.

$$x=\frac{4\sqrt{3}+4}{-2}$$

Divide $4+4\sqrt{3}$ by $-2$.

$$x=-2\sqrt{3}-2$$

Now solve the equation $x=\frac{4±4\sqrt{3}}{-2}$ when $±$ is minus. Subtract $4\sqrt{3}$ from $4$.

$$x=\frac{4-4\sqrt{3}}{-2}$$

Divide $4-4\sqrt{3}$ by $-2$.

$$x=2\sqrt{3}-2$$

The equation is now solved.

$$x=-2\sqrt{3}-2$$ $$x=2\sqrt{3}-2$$

Steps for Completing the Square

Add $x^{2}$ to both sides.

$$-2x^{2}+x+7+x^{2}=5x-1$$

Combine $-2x^{2}$ and $x^{2}$ to get $-x^{2}$.

$$-x^{2}+x+7=5x-1$$

Subtract $5x$ from both sides.

$$-x^{2}+x+7-5x=-1$$

Combine $x$ and $-5x$ to get $-4x$.

$$-x^{2}-4x+7=-1$$

Subtract $7$ from both sides.

$$-x^{2}-4x=-1-7$$

Subtract $7$ from $-1$ to get $-8$.

$$-x^{2}-4x=-8$$

Divide both sides by $-1$.

$$\frac{-x^{2}-4x}{-1}=-\frac{8}{-1}$$

Dividing by $-1$ undoes the multiplication by $-1$.

$$x^{2}+\left(-\frac{4}{-1}\right)x=-\frac{8}{-1}$$

Divide $-4$ by $-1$.

$$x^{2}+4x=-\frac{8}{-1}$$

Divide $-8$ by $-1$.

$$x^{2}+4x=8$$

Divide $4$, the coefficient of the $x$ term, by $2$ to get $2$. Then add the square of $2$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+4x+2^{2}=8+2^{2}$$

Square $2$.

$$x^{2}+4x+4=8+4$$

Add $8$ to $4$.

$$x^{2}+4x+4=12$$

Factor $x^{2}+4x+4$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+2\right)^{2}=12$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{12}$$

Simplify.

$$x+2=2\sqrt{3}$$ $$x+2=-2\sqrt{3}$$

Subtract $2$ from both sides of the equation.

$$x=2\sqrt{3}-2$$ $$x=-2\sqrt{3}-2$$