Solve Write the quadratic equation in standard form: 7x ^ 2 = - 19 - 5x + 5x ^ 2 | 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

$$\left. \begin{array} { l } { \quad 7 x ^ { 2 } = - 19 - 5 x + 5 x ^ { 2 } } \end{array} \right.$$

Solve for x

$x=\frac{-5+\sqrt{127}i}{4}\approx -1.25+2.817356917i$
$x=\frac{-\sqrt{127}i-5}{4}\approx -1.25-2.817356917i$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Subtract $-19$ from both sides.

$$7x^{2}-\left(-19\right)=-5x+5x^{2}$$

The opposite of $-19$ is $19$.

$$7x^{2}+19=-5x+5x^{2}$$

Add $5x$ to both sides.

$$7x^{2}+19+5x=5x^{2}$$

Subtract $5x^{2}$ from both sides.

$$7x^{2}+19+5x-5x^{2}=0$$

Combine $7x^{2}$ and $-5x^{2}$ to get $2x^{2}$.

$$2x^{2}+19+5x=0$$

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.

$$2x^{2}+5x+19=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $2$ for $a$, $5$ for $b$, and $19$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-5±\sqrt{5^{2}-4\times 2\times 19}}{2\times 2}$$

Square $5$.

$$x=\frac{-5±\sqrt{25-4\times 2\times 19}}{2\times 2}$$

Multiply $-4$ times $2$.

$$x=\frac{-5±\sqrt{25-8\times 19}}{2\times 2}$$

Multiply $-8$ times $19$.

$$x=\frac{-5±\sqrt{25-152}}{2\times 2}$$

Add $25$ to $-152$.

$$x=\frac{-5±\sqrt{-127}}{2\times 2}$$

Take the square root of $-127$.

$$x=\frac{-5±\sqrt{127}i}{2\times 2}$$

Multiply $2$ times $2$.

$$x=\frac{-5±\sqrt{127}i}{4}$$

Now solve the equation $x=\frac{-5±\sqrt{127}i}{4}$ when $±$ is plus. Add $-5$ to $i\sqrt{127}$.

$$x=\frac{-5+\sqrt{127}i}{4}$$

Now solve the equation $x=\frac{-5±\sqrt{127}i}{4}$ when $±$ is minus. Subtract $i\sqrt{127}$ from $-5$.

$$x=\frac{-\sqrt{127}i-5}{4}$$

The equation is now solved.

$$x=\frac{-5+\sqrt{127}i}{4}$$ $$x=\frac{-\sqrt{127}i-5}{4}$$

Steps for Completing the Square

Add $5x$ to both sides.

$$7x^{2}+5x=-19+5x^{2}$$

Subtract $5x^{2}$ from both sides.

$$7x^{2}+5x-5x^{2}=-19$$

Combine $7x^{2}$ and $-5x^{2}$ to get $2x^{2}$.

$$2x^{2}+5x=-19$$

Divide both sides by $2$.

$$\frac{2x^{2}+5x}{2}=-\frac{19}{2}$$

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

$$x^{2}+\frac{5}{2}x=-\frac{19}{2}$$

Divide $\frac{5}{2}$, the coefficient of the $x$ term, by $2$ to get $\frac{5}{4}$. Then add the square of $\frac{5}{4}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{2}x+\left(\frac{5}{4}\right)^{2}=-\frac{19}{2}+\left(\frac{5}{4}\right)^{2}$$

Square $\frac{5}{4}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=-\frac{19}{2}+\frac{25}{16}$$

Add $-\frac{19}{2}$ to $\frac{25}{16}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=-\frac{127}{16}$$

Factor $x^{2}+\frac{5}{2}x+\frac{25}{16}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+\frac{5}{4}\right)^{2}=-\frac{127}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{4}\right)^{2}}=\sqrt{-\frac{127}{16}}$$

Simplify.

$$x+\frac{5}{4}=\frac{\sqrt{127}i}{4}$$ $$x+\frac{5}{4}=-\frac{\sqrt{127}i}{4}$$

Subtract $\frac{5}{4}$ from both sides of the equation.

$$x=\frac{-5+\sqrt{127}i}{4}$$ $$x=\frac{-\sqrt{127}i-5}{4}$$