How to solve 14y=3+2(y)^(2)

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve 14y=3+2(y)^(2) | Equatix

Question

$$14y=3+2 { y }^{ 2 }$$

Solve for y

$y = \frac{\sqrt{43} + 7}{2} \approx 6.778719262$
$y=\frac{7-\sqrt{43}}{2}\approx 0.221280738$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Subtract $3$ from both sides.

$$14y-3=2y^{2}$$

Subtract $2y^{2}$ from both sides.

$$14y-3-2y^{2}=0$$

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.

$$-2y^{2}+14y-3=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-2$ for $a$, $14$ for $b$, and $-3$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$y=\frac{-14±\sqrt{14^{2}-4\left(-2\right)\left(-3\right)}}{2\left(-2\right)}$$

Square $14$.

$$y=\frac{-14±\sqrt{196-4\left(-2\right)\left(-3\right)}}{2\left(-2\right)}$$

Multiply $-4$ times $-2$.

$$y=\frac{-14±\sqrt{196+8\left(-3\right)}}{2\left(-2\right)}$$

Multiply $8$ times $-3$.

$$y=\frac{-14±\sqrt{196-24}}{2\left(-2\right)}$$

Add $196$ to $-24$.

$$y=\frac{-14±\sqrt{172}}{2\left(-2\right)}$$

Take the square root of $172$.

$$y=\frac{-14±2\sqrt{43}}{2\left(-2\right)}$$

Multiply $2$ times $-2$.

$$y=\frac{-14±2\sqrt{43}}{-4}$$

Now solve the equation $y=\frac{-14±2\sqrt{43}}{-4}$ when $±$ is plus. Add $-14$ to $2\sqrt{43}$.

$$y=\frac{2\sqrt{43}-14}{-4}$$

Divide $-14+2\sqrt{43}$ by $-4$.

$$y=\frac{7-\sqrt{43}}{2}$$

Now solve the equation $y=\frac{-14±2\sqrt{43}}{-4}$ when $±$ is minus. Subtract $2\sqrt{43}$ from $-14$.

$$y=\frac{-2\sqrt{43}-14}{-4}$$

Divide $-14-2\sqrt{43}$ by $-4$.

$$y=\frac{\sqrt{43}+7}{2}$$

The equation is now solved.

$$y=\frac{7-\sqrt{43}}{2}$$ $$y=\frac{\sqrt{43}+7}{2}$$

Steps for Completing the Square

Subtract $2y^{2}$ from both sides.

$$14y-2y^{2}=3$$

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

$$-2y^{2}+14y=3$$

Divide both sides by $-2$.

$$\frac{-2y^{2}+14y}{-2}=\frac{3}{-2}$$

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

$$y^{2}+\frac{14}{-2}y=\frac{3}{-2}$$

Divide $14$ by $-2$.

$$y^{2}-7y=\frac{3}{-2}$$

Divide $3$ by $-2$.

$$y^{2}-7y=-\frac{3}{2}$$

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

$$y^{2}-7y+\left(-\frac{7}{2}\right)^{2}=-\frac{3}{2}+\left(-\frac{7}{2}\right)^{2}$$

Square $-\frac{7}{2}$ by squaring both the numerator and the denominator of the fraction.

$$y^{2}-7y+\frac{49}{4}=-\frac{3}{2}+\frac{49}{4}$$

Add $-\frac{3}{2}$ to $\frac{49}{4}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}-7y+\frac{49}{4}=\frac{43}{4}$$

Factor $y^{2}-7y+\frac{49}{4}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(y-\frac{7}{2}\right)^{2}=\frac{43}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{7}{2}\right)^{2}}=\sqrt{\frac{43}{4}}$$

Simplify.

$$y-\frac{7}{2}=\frac{\sqrt{43}}{2}$$ $$y-\frac{7}{2}=-\frac{\sqrt{43}}{2}$$

Add $\frac{7}{2}$ to both sides of the equation.

$$y=\frac{\sqrt{43}+7}{2}$$ $$y=\frac{7-\sqrt{43}}{2}$$

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Example

Question

Solve for y

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Steps for Completing the Square