Solve (3x-5)/(2)-(4x+5)/(x)=(3)/(4)+x | 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

$$\frac{ 3x-5 }{ 2 } - \frac{ 4x+5 }{ x } = \frac{ 3 }{ 4 } +x$$

Solve for x

$x = \frac{\sqrt{1001} + 29}{4} \approx 15.15964601$
$x=\frac{29-\sqrt{1001}}{4}\approx -0.65964601$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $4x$, the least common multiple of $2,x,4$.

$$2x\left(3x-5\right)-4\left(4x+5\right)=4x\times \frac{3}{4}+4xx$$

Use the distributive property to multiply $2x$ by $3x-5$.

$$6x^{2}-10x-4\left(4x+5\right)=4x\times \frac{3}{4}+4xx$$

Use the distributive property to multiply $-4$ by $4x+5$.

$$6x^{2}-10x-16x-20=4x\times \frac{3}{4}+4xx$$

Combine $-10x$ and $-16x$ to get $-26x$.

$$6x^{2}-26x-20=4x\times \frac{3}{4}+4xx$$

Multiply $x$ and $x$ to get $x^{2}$.

$$6x^{2}-26x-20=4x\times \frac{3}{4}+4x^{2}$$

Cancel out $4$ and $4$.

$$6x^{2}-26x-20=3x+4x^{2}$$

Subtract $3x$ from both sides.

$$6x^{2}-26x-20-3x=4x^{2}$$

Combine $-26x$ and $-3x$ to get $-29x$.

$$6x^{2}-29x-20=4x^{2}$$

Subtract $4x^{2}$ from both sides.

$$6x^{2}-29x-20-4x^{2}=0$$

Combine $6x^{2}$ and $-4x^{2}$ to get $2x^{2}$.

$$2x^{2}-29x-20=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $2$ for $a$, $-29$ for $b$, and $-20$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-29\right)±\sqrt{\left(-29\right)^{2}-4\times 2\left(-20\right)}}{2\times 2}$$

Square $-29$.

$$x=\frac{-\left(-29\right)±\sqrt{841-4\times 2\left(-20\right)}}{2\times 2}$$

Multiply $-4$ times $2$.

$$x=\frac{-\left(-29\right)±\sqrt{841-8\left(-20\right)}}{2\times 2}$$

Multiply $-8$ times $-20$.

$$x=\frac{-\left(-29\right)±\sqrt{841+160}}{2\times 2}$$

Add $841$ to $160$.

$$x=\frac{-\left(-29\right)±\sqrt{1001}}{2\times 2}$$

The opposite of $-29$ is $29$.

$$x=\frac{29±\sqrt{1001}}{2\times 2}$$

Multiply $2$ times $2$.

$$x=\frac{29±\sqrt{1001}}{4}$$

Now solve the equation $x=\frac{29±\sqrt{1001}}{4}$ when $±$ is plus. Add $29$ to $\sqrt{1001}$.

$$x=\frac{\sqrt{1001}+29}{4}$$

Now solve the equation $x=\frac{29±\sqrt{1001}}{4}$ when $±$ is minus. Subtract $\sqrt{1001}$ from $29$.

$$x=\frac{29-\sqrt{1001}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{1001}+29}{4}$$ $$x=\frac{29-\sqrt{1001}}{4}$$

Steps for Completing the Square

Variable $x$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $4x$, the least common multiple of $2,x,4$.

$$2x\left(3x-5\right)-4\left(4x+5\right)=4x\times \frac{3}{4}+4xx$$

Use the distributive property to multiply $2x$ by $3x-5$.

$$6x^{2}-10x-4\left(4x+5\right)=4x\times \frac{3}{4}+4xx$$

Use the distributive property to multiply $-4$ by $4x+5$.

$$6x^{2}-10x-16x-20=4x\times \frac{3}{4}+4xx$$

Combine $-10x$ and $-16x$ to get $-26x$.

$$6x^{2}-26x-20=4x\times \frac{3}{4}+4xx$$

Multiply $x$ and $x$ to get $x^{2}$.

$$6x^{2}-26x-20=4x\times \frac{3}{4}+4x^{2}$$

Cancel out $4$ and $4$.

$$6x^{2}-26x-20=3x+4x^{2}$$

Subtract $3x$ from both sides.

$$6x^{2}-26x-20-3x=4x^{2}$$

Combine $-26x$ and $-3x$ to get $-29x$.

$$6x^{2}-29x-20=4x^{2}$$

Subtract $4x^{2}$ from both sides.

$$6x^{2}-29x-20-4x^{2}=0$$

Combine $6x^{2}$ and $-4x^{2}$ to get $2x^{2}$.

$$2x^{2}-29x-20=0$$

Add $20$ to both sides. Anything plus zero gives itself.

$$2x^{2}-29x=20$$

Divide both sides by $2$.

$$\frac{2x^{2}-29x}{2}=\frac{20}{2}$$

Dividing by $2$ undoes the multiplication by $2$.

$$x^{2}-\frac{29}{2}x=\frac{20}{2}$$

Divide $20$ by $2$.

$$x^{2}-\frac{29}{2}x=10$$

Divide $-\frac{29}{2}$, the coefficient of the $x$ term, by $2$ to get $-\frac{29}{4}$. Then add the square of $-\frac{29}{4}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{29}{2}x+\left(-\frac{29}{4}\right)^{2}=10+\left(-\frac{29}{4}\right)^{2}$$

Square $-\frac{29}{4}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{29}{2}x+\frac{841}{16}=10+\frac{841}{16}$$

Add $10$ to $\frac{841}{16}$.

$$x^{2}-\frac{29}{2}x+\frac{841}{16}=\frac{1001}{16}$$

Factor $x^{2}-\frac{29}{2}x+\frac{841}{16}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{29}{4}\right)^{2}=\frac{1001}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{29}{4}\right)^{2}}=\sqrt{\frac{1001}{16}}$$

Simplify.

$$x-\frac{29}{4}=\frac{\sqrt{1001}}{4}$$ $$x-\frac{29}{4}=-\frac{\sqrt{1001}}{4}$$

Add $\frac{29}{4}$ to both sides of the equation.

$$x=\frac{\sqrt{1001}+29}{4}$$ $$x=\frac{29-\sqrt{1001}}{4}$$