Question

In the following exercises, solve by using the Quadratic Formula.
$(y+4)(y-7)=18$

Short answer

The solutions of the given equation are $y=\frac{3}{2}\pm \frac{\sqrt{193}}{2}$.

Step-by-step solution

Step 1. Write the quadratic equation in standard form.

The given equation is:

$(y+4)(y-7)=18$

The given equation can be written as:

$\begin{array}{rcl}{y}^2-7y+4y-28& =& 18\\ y^2-3y-28-18& =& 0\\ y^2-3y-46& =& 0\end{array}$

Step 2. Substitute the values in Quadratic Formula.

Quadratic Formula for the equation $a{x}^2+bx+c=0$is:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

In the given equation, $a=1,b=-3,c=-46$. Substitute these values in Quadratic Formula.

$y=\frac{-(-3)\pm \sqrt{(-3{)}^2-4(1\left)\right(-46)}}{2\left(1\right)}$

Step 3. Simplify the value.

On simplification, we get

$\begin{array}{rcl}y& =& \frac{3\pm \sqrt{9+184}}{2}\\ y& =& \frac{3\pm \sqrt{193}}{2}\\ y& =& \frac{3}{2}\pm \frac{\sqrt{193}}{2}\end{array}$

Step 4. Conclusion.

The solutions of the given equation are $y=\frac{3}{2}\pm \frac{\sqrt{193}}{2}$.