Solve Write the quadratic equation in standard form: 6 - 8x = - 3x ^ 2 + 4 | 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

$$\left. \begin{array} { l } { \quad 6 - 8 x = - 3 x ^ { 2 } + 4 } \end{array} \right.$$

Solve for x

$x = \frac{\sqrt{10} + 4}{3} \approx 2.387425887$
$x=\frac{4-\sqrt{10}}{3}\approx 0.27924078$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Add $3x^{2}$ to both sides.

$$6-8x+3x^{2}=4$$

Subtract $4$ from both sides.

$$6-8x+3x^{2}-4=0$$

Subtract $4$ from $6$ to get $2$.

$$2-8x+3x^{2}=0$$

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.

$$3x^{2}-8x+2=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $3$ for $a$, $-8$ for $b$, and $2$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 3\times 2}}{2\times 3}$$

Square $-8$.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 3\times 2}}{2\times 3}$$

Multiply $-4$ times $3$.

$$x=\frac{-\left(-8\right)±\sqrt{64-12\times 2}}{2\times 3}$$

Multiply $-12$ times $2$.

$$x=\frac{-\left(-8\right)±\sqrt{64-24}}{2\times 3}$$

Add $64$ to $-24$.

$$x=\frac{-\left(-8\right)±\sqrt{40}}{2\times 3}$$

Take the square root of $40$.

$$x=\frac{-\left(-8\right)±2\sqrt{10}}{2\times 3}$$

The opposite of $-8$ is $8$.

$$x=\frac{8±2\sqrt{10}}{2\times 3}$$

Multiply $2$ times $3$.

$$x=\frac{8±2\sqrt{10}}{6}$$

Now solve the equation $x=\frac{8±2\sqrt{10}}{6}$ when $±$ is plus. Add $8$ to $2\sqrt{10}$.

$$x=\frac{2\sqrt{10}+8}{6}$$

Divide $8+2\sqrt{10}$ by $6$.

$$x=\frac{\sqrt{10}+4}{3}$$

Now solve the equation $x=\frac{8±2\sqrt{10}}{6}$ when $±$ is minus. Subtract $2\sqrt{10}$ from $8$.

$$x=\frac{8-2\sqrt{10}}{6}$$

Divide $8-2\sqrt{10}$ by $6$.

$$x=\frac{4-\sqrt{10}}{3}$$

The equation is now solved.

$$x=\frac{\sqrt{10}+4}{3}$$ $$x=\frac{4-\sqrt{10}}{3}$$

Steps for Completing the Square

Add $3x^{2}$ to both sides.

$$6-8x+3x^{2}=4$$

Subtract $6$ from both sides.

$$-8x+3x^{2}=4-6$$

Subtract $6$ from $4$ to get $-2$.

$$-8x+3x^{2}=-2$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form $x^{2}+bx=c$.

$$3x^{2}-8x=-2$$

Divide both sides by $3$.

$$\frac{3x^{2}-8x}{3}=-\frac{2}{3}$$

Dividing by $3$ undoes the multiplication by $3$.

$$x^{2}-\frac{8}{3}x=-\frac{2}{3}$$

Divide $-\frac{8}{3}$, the coefficient of the $x$ term, by $2$ to get $-\frac{4}{3}$. Then add the square of $-\frac{4}{3}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{8}{3}x+\left(-\frac{4}{3}\right)^{2}=-\frac{2}{3}+\left(-\frac{4}{3}\right)^{2}$$

Square $-\frac{4}{3}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{8}{3}x+\frac{16}{9}=-\frac{2}{3}+\frac{16}{9}$$

Add $-\frac{2}{3}$ to $\frac{16}{9}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{8}{3}x+\frac{16}{9}=\frac{10}{9}$$

Factor $x^{2}-\frac{8}{3}x+\frac{16}{9}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{4}{3}\right)^{2}=\frac{10}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{4}{3}\right)^{2}}=\sqrt{\frac{10}{9}}$$

Simplify.

$$x-\frac{4}{3}=\frac{\sqrt{10}}{3}$$ $$x-\frac{4}{3}=-\frac{\sqrt{10}}{3}$$

Add $\frac{4}{3}$ to both sides of the equation.

$$x=\frac{\sqrt{10}+4}{3}$$ $$x=\frac{4-\sqrt{10}}{3}$$