How do you solve $8x-{{x}^{2}}=15$ ?

Hint: In this question we have been asked to solve the given quadratic equation $8x-{{x}^{2}}=15$ . We know that we have a formula for finding the roots of quadratic expression in the form of $a{{x}^{2}}+bx+c=0$ which is given as $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .

Complete step by step answer:
Now considering from the question we have been asked to solve the given quadratic equation $8x-{{x}^{2}}=15$.
From the basic concepts of quadratic equations we know that we have a formula for finding the roots of quadratic expression in the form of $a{{x}^{2}}+bx+c=0$ which is given as $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .
Here in the given quadratic expression $8x-{{x}^{2}}=15\Rightarrow 15+{{x}^{2}}-8x=0$ the values of the variables $a,b,c$ are respectively $1,-8,15$ .
By applying the formula we will have
$\begin{align}
  & \Rightarrow \dfrac{8\pm \sqrt{{{\left( -8 \right)}^{2}}-4\left( 15 \right)}}{2}=\dfrac{8\pm \sqrt{64-60}}{2} \\
 & \Rightarrow \dfrac{8\pm \sqrt{4}}{2}=\dfrac{8\pm 2}{2} \\
 & \Rightarrow 5,3 \\
\end{align}$
So the roots of the given quadratic expression $8x-{{x}^{2}}=15$ are $3,5$

Therefore we can conclude that the solutions of the given quadratic expression $8x-{{x}^{2}}=15$ are 5,3.

Note: While answering questions of this type we should be sure with our concepts that we are going to apply during the process. This is a very simple question and can be solved accurately in a short span of time. We can also answer this question by the traditional method. The alternative method can be done by simplifying the given quadratic expression initially. By simplifying the expression we will have $-{{x}^{2}}+8x-15=0$ . Now we will write the coefficient of $x$ as the sum of the factors of the product of the constant and the coefficient of ${{x}^{2}}$ . This implies that the coefficient of $x$ can be written using $-1\times -15=15\Rightarrow 3\times 5$ . So now we will have
 $\begin{align}
  & \Rightarrow -{{x}^{2}}+3x+5x-15=0 \\
 & \Rightarrow -x\left( x-3 \right)+5\left( x-3 \right)=0 \\
 & \Rightarrow \left( x-3 \right)\left( -x+5 \right)=0 \\
\end{align}$
Hence the solutions are $3,5$ .