Solve (5 - x) = (x ^ 2 * 25)/7 | Equatix - Math Solver

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Example

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

a⋅(b-2)=3b

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

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Question

$$(5-x)=\frac{x^{2}\cdot25}{7}$$

Solve for x

$x = \frac{13 \sqrt{21} - 7}{50} \approx 1.051469681$
$x=\frac{-13\sqrt{21}-7}{50}\approx -1.331469681$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Multiply both sides of the equation by $7$.

$$35-7x=x^{2}\times 25$$

Subtract $x^{2}\times 25$ from both sides.

$$35-7x-x^{2}\times 25=0$$

Multiply $-1$ and $25$ to get $-25$.

$$35-7x-25x^{2}=0$$

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.

$$-25x^{2}-7x+35=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-25$ for $a$, $-7$ for $b$, and $35$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\left(-25\right)\times 35}}{2\left(-25\right)}$$

Square $-7$.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\left(-25\right)\times 35}}{2\left(-25\right)}$$

Multiply $-4$ times $-25$.

$$x=\frac{-\left(-7\right)±\sqrt{49+100\times 35}}{2\left(-25\right)}$$

Multiply $100$ times $35$.

$$x=\frac{-\left(-7\right)±\sqrt{49+3500}}{2\left(-25\right)}$$

Add $49$ to $3500$.

$$x=\frac{-\left(-7\right)±\sqrt{3549}}{2\left(-25\right)}$$

Take the square root of $3549$.

$$x=\frac{-\left(-7\right)±13\sqrt{21}}{2\left(-25\right)}$$

The opposite of $-7$ is $7$.

$$x=\frac{7±13\sqrt{21}}{2\left(-25\right)}$$

Multiply $2$ times $-25$.

$$x=\frac{7±13\sqrt{21}}{-50}$$

Now solve the equation $x=\frac{7±13\sqrt{21}}{-50}$ when $±$ is plus. Add $7$ to $13\sqrt{21}$.

$$x=\frac{13\sqrt{21}+7}{-50}$$

Divide $7+13\sqrt{21}$ by $-50$.

$$x=\frac{-13\sqrt{21}-7}{50}$$

Now solve the equation $x=\frac{7±13\sqrt{21}}{-50}$ when $±$ is minus. Subtract $13\sqrt{21}$ from $7$.

$$x=\frac{7-13\sqrt{21}}{-50}$$

Divide $7-13\sqrt{21}$ by $-50$.

$$x=\frac{13\sqrt{21}-7}{50}$$

The equation is now solved.

$$x=\frac{-13\sqrt{21}-7}{50}$$ $$x=\frac{13\sqrt{21}-7}{50}$$

Steps for Completing the Square

Multiply both sides of the equation by $7$.

$$35-7x=x^{2}\times 25$$

Subtract $x^{2}\times 25$ from both sides.

$$35-7x-x^{2}\times 25=0$$

Multiply $-1$ and $25$ to get $-25$.

$$35-7x-25x^{2}=0$$

Subtract $35$ from both sides. Anything subtracted from zero gives its negation.

$$-7x-25x^{2}=-35$$

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 $x^{2}+bx=c$.

$$-25x^{2}-7x=-35$$

Divide both sides by $-25$.

$$\frac{-25x^{2}-7x}{-25}=-\frac{35}{-25}$$

Dividing by $-25$ undoes the multiplication by $-25$.

$$x^{2}+\left(-\frac{7}{-25}\right)x=-\frac{35}{-25}$$

Divide $-7$ by $-25$.

$$x^{2}+\frac{7}{25}x=-\frac{35}{-25}$$

Reduce the fraction $\frac{-35}{-25}$ to lowest terms by extracting and canceling out $5$.

$$x^{2}+\frac{7}{25}x=\frac{7}{5}$$

Divide $\frac{7}{25}$, the coefficient of the $x$ term, by $2$ to get $\frac{7}{50}$. Then add the square of $\frac{7}{50}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{7}{25}x+\left(\frac{7}{50}\right)^{2}=\frac{7}{5}+\left(\frac{7}{50}\right)^{2}$$

Square $\frac{7}{50}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{7}{25}x+\frac{49}{2500}=\frac{7}{5}+\frac{49}{2500}$$

Add $\frac{7}{5}$ to $\frac{49}{2500}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{7}{25}x+\frac{49}{2500}=\frac{3549}{2500}$$

Factor $x^{2}+\frac{7}{25}x+\frac{49}{2500}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+\frac{7}{50}\right)^{2}=\frac{3549}{2500}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{50}\right)^{2}}=\sqrt{\frac{3549}{2500}}$$

Simplify.

$$x+\frac{7}{50}=\frac{13\sqrt{21}}{50}$$ $$x+\frac{7}{50}=-\frac{13\sqrt{21}}{50}$$

Subtract $\frac{7}{50}$ from both sides of the equation.

$$x=\frac{13\sqrt{21}-7}{50}$$ $$x=\frac{-13\sqrt{21}-7}{50}$$