Solve (x ^ 2 - 5x)/2 + x = 15 | 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

$$\frac{x^{2}-5x}{2}+x=15.$$

Solve for x

$x = \frac{\sqrt{129} + 3}{2} \approx 7.178908346$
$x=\frac{3-\sqrt{129}}{2}\approx -4.178908346$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Multiply both sides of the equation by $2$.

$$x^{2}-5x+2x=30$$

Combine $-5x$ and $2x$ to get $-3x$.

$$x^{2}-3x=30$$

Subtract $30$ from both sides.

$$x^{2}-3x-30=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $-3$ for $b$, and $-30$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-30\right)}}{2}$$

Square $-3$.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-30\right)}}{2}$$

Multiply $-4$ times $-30$.

$$x=\frac{-\left(-3\right)±\sqrt{9+120}}{2}$$

Add $9$ to $120$.

$$x=\frac{-\left(-3\right)±\sqrt{129}}{2}$$

The opposite of $-3$ is $3$.

$$x=\frac{3±\sqrt{129}}{2}$$

Now solve the equation $x=\frac{3±\sqrt{129}}{2}$ when $±$ is plus. Add $3$ to $\sqrt{129}$.

$$x=\frac{\sqrt{129}+3}{2}$$

Now solve the equation $x=\frac{3±\sqrt{129}}{2}$ when $±$ is minus. Subtract $\sqrt{129}$ from $3$.

$$x=\frac{3-\sqrt{129}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{129}+3}{2}$$ $$x=\frac{3-\sqrt{129}}{2}$$

Steps for Completing the Square

Multiply both sides of the equation by $2$.

$$x^{2}-5x+2x=30$$

Combine $-5x$ and $2x$ to get $-3x$.

$$x^{2}-3x=30$$

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

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=30+\left(-\frac{3}{2}\right)^{2}$$

Square $-\frac{3}{2}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=30+\frac{9}{4}$$

Add $30$ to $\frac{9}{4}$.

$$x^{2}-3x+\frac{9}{4}=\frac{129}{4}$$

Factor $x^{2}-3x+\frac{9}{4}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{129}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{129}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{\sqrt{129}}{2}$$ $$x-\frac{3}{2}=-\frac{\sqrt{129}}{2}$$

Add $\frac{3}{2}$ to both sides of the equation.

$$x=\frac{\sqrt{129}+3}{2}$$ $$x=\frac{3-\sqrt{129}}{2}$$