Solve Write the quadratic equation in standard form: 4x + 17 + 5x ^ 2 = - 3x | 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

$$\left. \begin{array} { l } { \quad 4 x + 17 + 5 x ^ { 2 } = - 3 x } \end{array} \right.$$

Solve for x

$x=\frac{-7+\sqrt{291}i}{10}\approx -0.7+1.705872211i$
$x=\frac{-\sqrt{291}i-7}{10}\approx -0.7-1.705872211i$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Add $3x$ to both sides.

$$4x+17+5x^{2}+3x=0$$

Combine $4x$ and $3x$ to get $7x$.

$$7x+17+5x^{2}=0$$

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.

$$5x^{2}+7x+17=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $5$ for $a$, $7$ for $b$, and $17$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-7±\sqrt{7^{2}-4\times 5\times 17}}{2\times 5}$$

Square $7$.

$$x=\frac{-7±\sqrt{49-4\times 5\times 17}}{2\times 5}$$

Multiply $-4$ times $5$.

$$x=\frac{-7±\sqrt{49-20\times 17}}{2\times 5}$$

Multiply $-20$ times $17$.

$$x=\frac{-7±\sqrt{49-340}}{2\times 5}$$

Add $49$ to $-340$.

$$x=\frac{-7±\sqrt{-291}}{2\times 5}$$

Take the square root of $-291$.

$$x=\frac{-7±\sqrt{291}i}{2\times 5}$$

Multiply $2$ times $5$.

$$x=\frac{-7±\sqrt{291}i}{10}$$

Now solve the equation $x=\frac{-7±\sqrt{291}i}{10}$ when $±$ is plus. Add $-7$ to $i\sqrt{291}$.

$$x=\frac{-7+\sqrt{291}i}{10}$$

Now solve the equation $x=\frac{-7±\sqrt{291}i}{10}$ when $±$ is minus. Subtract $i\sqrt{291}$ from $-7$.

$$x=\frac{-\sqrt{291}i-7}{10}$$

The equation is now solved.

$$x=\frac{-7+\sqrt{291}i}{10}$$ $$x=\frac{-\sqrt{291}i-7}{10}$$

Steps for Completing the Square

Add $3x$ to both sides.

$$4x+17+5x^{2}+3x=0$$

Combine $4x$ and $3x$ to get $7x$.

$$7x+17+5x^{2}=0$$

Subtract $17$ from both sides. Anything subtracted from zero gives its negation.

$$7x+5x^{2}=-17$$

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

$$5x^{2}+7x=-17$$

Divide both sides by $5$.

$$\frac{5x^{2}+7x}{5}=-\frac{17}{5}$$

Dividing by $5$ undoes the multiplication by $5$.

$$x^{2}+\frac{7}{5}x=-\frac{17}{5}$$

Divide $\frac{7}{5}$, the coefficient of the $x$ term, by $2$ to get $\frac{7}{10}$. Then add the square of $\frac{7}{10}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{7}{5}x+\left(\frac{7}{10}\right)^{2}=-\frac{17}{5}+\left(\frac{7}{10}\right)^{2}$$

Square $\frac{7}{10}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{7}{5}x+\frac{49}{100}=-\frac{17}{5}+\frac{49}{100}$$

Add $-\frac{17}{5}$ to $\frac{49}{100}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{7}{5}x+\frac{49}{100}=-\frac{291}{100}$$

Factor $x^{2}+\frac{7}{5}x+\frac{49}{100}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+\frac{7}{10}\right)^{2}=-\frac{291}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{10}\right)^{2}}=\sqrt{-\frac{291}{100}}$$

Simplify.

$$x+\frac{7}{10}=\frac{\sqrt{291}i}{10}$$ $$x+\frac{7}{10}=-\frac{\sqrt{291}i}{10}$$

Subtract $\frac{7}{10}$ from both sides of the equation.

$$x=\frac{-7+\sqrt{291}i}{10}$$ $$x=\frac{-\sqrt{291}i-7}{10}$$